ADDISON
WESLEY
■■■
MECHANICS
Second Edition
By Keith i;. Symon
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THE AUTHOR
Keith H Sya boS.B., A M ,etnd
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,,1 I During 194.J 945 Profesaoi
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Lchitig duties p ■ er y ■ Wisi nnsin,
ia ;,i p ■-. !• -.: '■■ the Midwestern Uni-
• 1 1 -M i-.-- Resea rel A.----"i- i : m
dffirgn .,: -i. ;■ ilti I'-- -•■<> tide ■!'■■ derator Hia
other fields d ii - ave included design
ultra-high frequency radio equipment, the i
of eoarg ona if Fasl particles,
-i ttistical communicai on theory und classici I
Lheqn ticaJ phya ca
1 ; : p
Mechanics
Symm
I
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MECHANICS
This book is in the
ADDISON-WESLEY SERIES IN PHYSICS
MECHANICS
by
KEITH R. SYMON
University of Wisconsin
SECOND EDITION
ADDISON-WESLEY PUBLISHING COMPANY, INC.
BEADING, MASSACHUSETTS, U.S.A.
LONDON, ENGLAND
Copyright © 1953, 1960
ADDISON-WESLEY PUBLISHING COMPANY, INC.
Printed in the United States of America
ALL EIGHTS EESEBVED. THIS BOOK, OR PARTS THERE-
OF, MAY NOT BE REPRODUCED IN ANY FORM WITH-
OUT WRITTEN PERMISSION OF THE PUBLISHER.
Library of Congress Catalog Card No. 60-5164
Second Edition
To my Father
PREFACE
This text is intended as the basis for an intermediate course in mechanics
at the undergraduate level. Such a course, as essential preparation for
advanced work in physics, has several major objectives. It must develop
in the student a thorough understanding of the fundamental principles of
mechanics. It should treat in detail certain specific problems of primary
importance in physics, for example, the harmonic oscillator, and the
motion of a particle under a central force. The problems suggested and
those worked out in the text have been chosen with regard to their in-
terest and importance in physics, as well as to their instructive value.
This book contains sufficient material for a two-semester course, and is
arranged in such a way that, with appropriate omissions, it can be used
for a single three- or four-hour course for one semester. The author has
used the material in the first seven chapters in a three-hour course in
mechanics.
The choice of topics and their treatment throughout the book are in-
tended to emphasize the modern point of view. Applications to atomic
physics are made wherever possible, with an indication as to the extent of
the validity of the results of classical mechanics. The inadequacies in
classical mechanics are carefully pointed out, and the points of departure
for quantum mechanics and for relativistic mechanics are indicated. The
development, except for the last four chapters, proceeds directly from
Newton's laws of motion, which form a suitable basis from which to attack
most mechanical problems. More advanced methods, using Lagrange's
equations and tensor algebra, are introduced in the last four chapters.
An important objective of a first course in mechanics is to train the
student to think about physical phenomena in mathematical terms. Most
students have a fairly good intuitive feeling for mechanical phenomena in
a qualitative way. The study of mechanics should aim at developing an
almost equally intuitive feeling for the precise mathematical formulation
of physical problems and for the physical interpretation of the mathe-
matical solutions. The examples treated in the text have been worked
out so as to integrate, as far as possible, the mathematical treatment with
the physical interpretation. After working an assigned problem, the
student should study it until he is sure he understands the physical inter-
pretation of every feature of the mathematical treatment. He should de-
cide whether the result agrees with his physical intuition about the prob-
lem. If not, then either his solution or his intuition should be appropriately
corrected. If the answer is fairly complicated, he should try to see whether
vii
Vlll PREFACE
it can be simplified in certain special or limiting cases. He should try to
formulate and solve similar problems on his own.
Only a knowledge of differential and integral calculus has been presup-
posed. Mathematical concepts beyond those treated in the first year of
calculus are introduced and explained as needed. A previous course in
elementary differential equations or vector analysis may be helpful, but it
is the author's experience that students with an adequate preparation in
algebra and calculus are able to handle the vector analysis and differential
equations needed for this course with the explanations provided herein.
A physics student is likely to get more out of his advanced courses in
mathematics if he has previously encountered these concepts in physics.
The text has been written so as to afford maximum flexibility in the
selection and arrangement of topics to be covered. With certain obvious
exceptions, many sections or groups of sections can be postponed or
omitted without prejudice to the understanding of the remaining material.
Where particular topics presented earlier are needed in later parts of the
book, references to section and equation numbers make it easy to locate
the earlier material needed.
In the first chapter the basic concepts of mechanics are reviewed, and
the laws of mechanics and of gravitation are formulated and applied to a
few simple examples. The second chapter undertakes a fairly thorough
study of the problem of one-dimensional motion. The chapter concludes
with a study of the harmonic oscillator as probably the most important
example of one-dimensional motion. Use is made of complex numbers to
represent oscillating quantities. The last section, on the principle of super-
position, makes some use of Fourier series, and provides a basis for certain
parts of Chapters 8 and 12. If these chapters are not to be covered, Sec-
tion 2-11 may be omitted Or, better, skimmed to get a brief indication of
the significance of the principle of superposition and the way in which
Fourier series are used to treat the problem of an arbitrary applied force
function.
Chapter 3 begins with a development of vector algebra and its use in
describing motions in a plane or in space. Boldface letters are used for
vectors. Section 3-6 is a brief introduction to vector analysis, which is
used very little in this book except in Chapter 8, and it may be omitted
or skimmed if Chapter 8 and a few proofs in some other chapters are
omitted. The author feels there is some advantage in introducing the
student to the concepts and notation of vector analysis at this stage, where
the level of treatment is fairly easy; in later courses where the physical
concepts and mathematical treatment become more difficult, it will be well
if the notations are already familiar. The theorems stating the time rates
of change of momentum, energy, and angular momentum are derived for
a moving particle, and several problems are discussed, of which motion
PREFACK ix
under central forces receives major attention. Examples are taken from
astronomical and from atomic problems.
In Chapter 4 the conservation laws of energy, momentum, and angular
momentum are derived, with emphasis on their position as cornerstones of
present-day physics. They are then applied to typical problems, particu-
larly collision problems. The two-body problem is solved, and the motion
of two coupled harmonic oscillators is worked out. The general theory of
coupled oscillations is best treated by means of linear transformations in
vector spaces, as in Chapter 12, but the behavior of coupled oscillating
systems is too important to be omitted altogether from even a one-semester
course. The section on two coupled oscillators can be omitted or postponed
until Chapter 12. The rigid body is discussed in Chapter 5 as a special
kind of system of particles. Only rotation about a fixed axis is treated;
the more general study of the motion of a rigid body is left to a later chap-
ter, where more advanced methods are used. The section on statics treats
the problem of the reduction of a system of forces to an equivalent simpler
system. Elementary treatments of the equilibrium of beams, flexible
strings, and of fluids are given in Sections 5-9, 5-10, and 5-11.
The theory of gravitation is studied in some detail in Chapter 6. The
last section, on the gravitational field equations, may be omitted without
disturbing the continuity of the remaining material. The laws of motion
in moving coordinate systems are worked out in Chapter 7, and applied
to motion on the rotating earth and to the motion of a system of charged
particles in a magnetic field. Particular attention is paid to the status
in Newtonian mechanics of the "fictitious forces" which appear when
moving coordinate systems are introduced, and to the role to be played
by such forces in the general theory of relativity.
The last five chapters cover more advanced material and are designed
primarily to be used in the second semester of a two-semester course in
intermediate mechanics. In a shorter course, any or all of the last five
chapters may be omitted without destroying the unity of the course,
although the author has found it possible to utilize parts of Chapter 8 or
9 even in a one-semester course. In Chapter 8 an introductory treatment
of vibrating strings and of the motion of fluids is presented, with emphasis
on the fundamental concepts and mathematical methods used in treating
the mechanics of continuous media. Chapter 9 on Lagrange's equations
is intended as an introduction to the methods of advanced dynamics.
Hamilton's equations and the concept of phase space are presented, since
they are prerequisite to any later course in quantum mechanics or statis-
tical mechanics, but the theory of canonical transformations and the use
of variational principles are beyond the scope of this book. Chapter 10
develops the algebra of tensors, including orthogonal coordinate trans-
formations, which are required in the last two chapters. The inertia tensor
X PREFACE
and the stress tensor are described in some detail as examples. Section
10-6 on the stress tensor will enable the reader to extend the discussion
of ideal fluids in Chapter 8 to a solid or viscous medium. The methods
developed in Chapters 9 and 10 are applied in Chapter 11 to the general
rotation of a rigid body about a point, and in Chapter 12 to the study of
small vibrations of a physical system about a state of equilibrium or of
steady motion.
The problems at the end of each chapter are arranged in the order
in which the material is covered in the chapter, for convenience in as-
signment. An attempt has been made to include a sufficient variety of
problems to guarantee that anyone who can solve them has mastered the
material in the text. The converse is not necessarily true, since most
problems require more or less physical ingenuity in addition to an under-
standing of the text. Many of the problems are fairly easy and should be
tractable for anyone who has understood the material presented. A few
are probably too difficult for most college juniors or seniors to solve with-
out some assistance. Those problems which are particularly difficult or
time-consuming are marked with an asterisk.
The last three chapters and the last three sections of Chapter 9 have
been added to the present edition in order to provide enough material for
a full two-semester course in mechanics. Except for corrections and a
few minor changes and additions, the first eight chapters and the first
eight sections of Chapter 9 remain the same as in the first edition of this
text.
Grateful acknowledgment is made to Professor Francis W. Sears of
Dartmouth College and to Professor George H. Vineyard of Brookhaven
National Laboratory for their many helpful suggestions, and to Mr. Charles
Vittitoe and Mr. Donald Roiseland for a critical reading of the last four
chapters. The author is particularly grateful to the many teachers and
students who have offered corrections and suggestions for improvement
which have been incorporated in this revised edition. While space does not
permit mentioning individuals here, I hope that each may find my thanks
expressed in the changes that have been made in this edition.
January, 1960 K. R. S.
CONTENTS
Chapter 1. Elements of Newtonian Mechanics 1
1-1 Mechanics, an exact science 1
1-2 Kinematics, the description of motion 4
1-3 Dynamics. Mass and force 5
1-4 Newton's laws of motion 7
1-5 Gravitation 10
1-6 Units and dimensions 11
1-7 Some elementary problems in mechanics 13
Chapter 2. Motion of a Particle in One Dimension .... 21
2-1 Momentum and energy theorems 21
2-2 Discussion of the general problem of one-dimensional motion . 22
2-3 Applied force depending on the time 25
2-4 Damping force depending on the velocity 28
2-5 Conservative force depending on position. Potential energy . 30
2-6 Falling bodies 35
2-7 The simple harmonic oscillator 39
2-8 Linear differential equations with constant coefficients ... 41
2-9 The damped harmonic oscillator 47
2-10 The forced harmonic oscillator 50
2-11 The principle of superposition. Harmonic oscillator with arbitrary
applied force 59
Chapter 3. Motion of a Particle in Two ok Three Dimensions 68
3-1 Vector algebra 68
3-2 Applications to a set of forces acting on a particle .... 77
3-3 Differentiation and integration of vectors 81
3-4 Kinematics in a plane 87
3-5 Kinematics in three dimensions 91
3-6 Elements of vector analysis 95
3-7 Momentum and energy theorems 100
3-8 Plane and vector angular momentum theorems 101
3-9 Discussion of the general problem of two- and three-dimensional
motion 104
3-10 The harmonic oscillator in two and three dimensions .... 106
3-11 Projectiles 108
3-12 Potential energy . 112
3-13 Motion under a central force 120
3-14 The central force inversely proportional to the square of the
distance 125
xi
Xll CONTENTS
3-15 Elliptic orbits. The Kepler problem 132
3-16 Hyperbolic orbits. The Rutherford problem. Scattering cross
section 135
3-17 Motion of a particle in an electromagnetic field 139
Chapter 4. The Motion of a System of Particles 155
4-1 Conservation of linear momentum. Center of mass .... 155
4-2 Conservation of angular momentum 158
4-3 Conservation of energy 162
4-4 Critique of the conservation laws 165
4-5 Rockets, conveyor belts, and planets 168
4-6 Collision problems 171
4-7 The two-body problem 178
4-8 Center-of-mass coordinates. Rutherford scattering by a charged
particle of finite mass 181
4-9 The IV-body problem 185
4-10 Two coupled harmonic oscillators 188
Chapter 5. Rigid Bodies. Rotation About an Axis. Statics . . 203
5-1 The dynamical problem of the motion of a rigid body . . . 203
5-2 Rotation about an axis 206
5-3 The simple pendulum 208
5-4 The compound pendulum 212
5-5 Computation of centers of mass and moments of inertia . . . 215
5-6 Statics of rigid bodies 225
5-7 Statics of structures 231
5-8 Stress and strain 232
5-9 Equilibrium of flexible strings and cables 235
5-10 Equilibrium of solid beams 239
5-11 Equilibrium of fluids 245
Chapter 6. Gravitation 257
6-1 Centers of gravity for extended bodies 257
6-2 Gravitational field and gravitational potential 259
6-3 Gravitational field equations 262
Chapter 7. Moving Coordinate Systems 269
7-1 Moving origin of coordinates 269
7-2 Rotating coordinate systems 271
7-3 Laws of motion on the rotating earth 278
7—4 The Foucault pendulum 280
7-5 Larmor's theorem 283
7-6 The restricted three-body problem 285
CONTENTS Xiii
Chapter 8. Introduction to the Mechanics of Continuous Media 294
8-1 The equation of motion for the vibrating string . . . . . . 294
8-2 Normal modes of vibration for the vibrating string .... 296
8-3 Wave propagation along a string 300
8-4 The string as a limiting case of a system of particles . . , . 305
8-5 General remarks on the propagation of waves ...... 310
8-6 Kinematics of moving fluids 313
8-7 Equations of motion for an ideal fluid 321
8-8 Conservation laws for fluid motion 323
8-9 Steady flow 329
8-10 Sound waves 332
8-11 Normal vibrations of fluid in a rectangular box 337
8-12 Sound waves in pipes 341
8-13 The Mach number 343
8-14 Viscosity 345
Chapter 9. Lagrange's Equations 354
9-1 Generalized coordinates 354
9-2 Lagrange's equations . . 365
9-3 Examples 3gg
9-4 Systems subject to constraints 369
9-5 Examples of systems subject to constraints 375
9-6 Constants of the motion and ignorable coordinates .... 381
9-7 Further examples 384
9-8 Electromagnetic forces and velocity-dependent potentials . . 388
9-9 Lagrange's equations for the vibrating string 391
9-10 Hamilton's equations 396
9-11 Liouville's theorem 399
Chapter 10. Tensor Algebra. Inertia and Stress Tensors . . 406
10-1 Angular momentum of a rigid body . 406
10-2 Tensor algebra 407
10-3 Coordinate transformations 414
10-4 Diagonalization of a symmetric tensor 421
10-5 The inertia tensor 430
10-6 The stress tensor 438
Chapter 11. The Rotation of a Rigid Body 450
11-1 Motion of a rigid body in space 450
11-2 Euler's equations of motion for a rigid body 451
1 1-3 Poinsot's solution for a freely rotating body 455
11-4 Euler's angles 458
11-5 The symmetrical top 461
Xiv CONTENTS
Chapteb 12. Theokt of Small Vibrations 473
12-1 Condition for stability near an equilibrium configuration . 473
12-2 Linearized equations of motion near an equilibrium configuration 475
12-3 Normal modes of vibration 477
12-4 Forced vibrations 481
12-5 Perturbation theory 484
12-6 Small vibrations about steady motion 490
12-7 Betatron oscillations in an accelerator 497
12-8 Stability of Lagrange's three bodies 500
Bibliography 515
Answers to Odd-Numbered Problems 521
List of Symbols 533
Index 545
CHAPTER 1
ELEMENTS OF NEWTONIAN MECHANICS
1-1 Mechanics, an exact science. When we say that physics is an
exact science, we mean that its laws are expressed in the form of mathe-
matical equations which describe and predict the results of precise quanti-
tative measurements. The advantage in a quantitative physical theory is
not alone the practical one that it gives us the power accurately to predict
and to control natural phenomena. By a comparison of the results of
accurate measurements with the numerical predictions of the theory, we
can gain considerable confidence that the theory is correct, and we can
determine in what respects it needs to be modified. It is often possible
to explain a given phenomenon in several rough qualitative ways, and if
we are content with that, it may be impossible to decide which theory is
correct. But if a theory can be given which predicts correctly the results
of measurements to four or five (or even two or three) significant figures,
the theory can hardly be very far wrong. Rough agreement might be a
coincidence, but close agreement is unlikely to be. Furthermore, there
have been many cases in the history of science when small but significant
discrepancies between theory and accurate measurements have led to the
development of new and more far-reaching theories. Such slight discrep-
ancies would not even have been detected if we had been content with a
merely qualitative explanation of the phenomena.
The symbols which are to appear in the equations that express the laws
of a science must represent quantities which can be expressed in numerical
terms. Hence the concepts in terms of which an exact science is to be
developed must be given precise numerical meanings. If a definition of a
quantity (mass, for example) is to be given, the definition must be such
as to specify precisely how the value of the quantity is to be determined
in any given case. A qualitative remark about its meaning may be helpful,
but is not sufficient as a definition. As a matter of fact, it is probably not
possible to give an ideally precise definition of every concept appearing in
a physical theory. Nevertheless, when we write down a mathematical
equation, the presumption is that the symbols appearing in it have precise
meanings, and we should strive to make our ideas as clear and precise as
possible, and to recognize at what points there is a lack of precision or
clarity. Sometimes a new concept can be defined in terms of others whose
meanings are known, in which case there is no problem. For example,
momentum = mass X velocity
1
2 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1
gives a perfectly precise definition of "momentum" provided "mass" and
"velocity" are assumed to be precisely defined already. But this kind of
definition will not do for all terms in a theory, since we must start some-
where with a set of basic concepts or "primitive" terms whose meanings
are assumed known. The first concepts to be introduced in a theory can-
not be defined in the above way, since at first we have nothing to put on
the right side of the equation. The meanings of these primitive terms
must be made clear by some means that lies outside of the physical theories
being set up. We might, for example, simply use the terms over and over
until their meanings become clear. This is the way babies learn a language,
and probably, to some extent, freshman physics students learn the same
way. We might define all primitive terms by stating their meaning in
terms of observation and experiment. In particular, nouns designating
measurable quantities, like force, mass, etc., may be defined by specifying
the operational process for measuring them. One school of thought holds
that all physical terms should be defined in this way. Or we might simply
state what the primitive terms are, with a rough indication of their physi-
cal meaning, and then let the meaning be determined more precisely by
the laws and postulates we lay down and the rules that we give for inter-
preting theoretical results in terms of experimental situations. This is the
most convenient and flexible way, and is the way physical theories are
usually set up. It has the disadvantage that we are never sure that our
concepts have been given a precise meaning. It is left to experience to
decide not only whether our laws are correct, but even whether the con-
cepts we use have a precise meaning. The modern theories of relativity
and quanta arise as much from fuzziness in classical concepts as from in-
accuracies in classical laws.
Historically, mechanics was the earliest branch of physics to be developed
as an exact science. The laws of levers and of fluids in static equilibrium
were known to Greek scientists in the third century B.C. The tremendous
development of physics in the last three centuries began with the discovery
of the laws qf mechanics by Galileo and Newton. The laws of mechanics
as formulated by Isaac Newton in the middle of the seventeenth century
and the laws of electricity and magnetism as formulated by James Clerk
Maxwell about two hundred years later are the two basic theories of classi-
cal physics. Relativistic physics, which began with the work of Einstein
in 1905, and quantum physics, as based upon the work of Heisenberg
and Schroedinger in 1925-1926, require a modification and reformulation
of mechanics and electrodynamics in terms of new physical concepts.
Nevertheless, modern physics builds on the foundations laid by classical
physics, and a clear understanding of the principles of classical mechanics
and electrodynamics is still essential in the study of relativistic and quan-
tum physics. Furthermore, in the vast majority of practical applications
of mechanics to the various branches of engineering and to astronomy, the
1-1] MECHANICS, AN EXACT SCIENCE 3
laws of classical mechanics can still be applied. Except when bodies travel
at speeds approaching the speed of light, or when enormous masses or
enormous distances are involved, relativistic mechanics gives the same re-
sults as classical mechanics; indeed, it must, since we know from experi-
ence that classical mechanics gives correct results in ordinary applications.
Similarly, quantum mechanics should and does agree with classical mechan-
ics except when applied to physical systems of molecular size or smaller.
Indeed, one of the chief guiding principles in formulating new physical
theories is the requirement that they must agree with the older theories
when applied to those phenomena where the older theories are known to
be correct.
Mechanics is the study of the motions of material bodies. Mechanics
may be divided into three subdisciplines, kinematics, dynamics, and statics.
Kinematics is the study and description of the possible motions of mate-
rial bodies. Dynamics is the study of the laws which determine, among
all possible motions, which motion will actually take place in any given
case. In dynamics we introduce the concept of force. The central prob-
lem of dynamics is to determine for any physical system the motions which
will take place under the action of given forces. Statics is the study of
forces and systems of forces, with particular reference to systems of forces
which act on bodies at rest.
We may also subdivide the study of mechanics according to the kind of
physical system to be studied. This is, in general, the basis for the outline
of the present book. The simplest physical system, and the one we shall
study first, is a single particle. Next we shall study the motion of a sys-
tem of particles. A rigid body may be treated as a special kind of system
of particles. Finally, we shall study the motions of continuous media,
elastic and plastic substances, solids, liquids, and gases.
A great many of the applications of classical mechanics may be based
directly on Newton's laws of motion. All of the problems studied in this
book, except in Chapters 9-12, are treated in this way. There are, how-
ever, a number of other ways of formulating the principles of classical
mechanics. The equations of Lagrange and of Hamilton are examples.
They are not new physical theories, for they may be derived from Newton's
laws, but they are different ways of expressing the same physical theory.
They use more advanced mathematical concepts, they are in some respects
more elegant than Newton's formulation, and they are in some cases more
powerful in that they allow the solutions of some problems whose solution
based directly on Newton's laws would be very difficult. The more differ-
ent ways we know to formulate a physical theory, the better chance we
have of learning how to modify it to fit new kinds of phenomena as they
are discovered. This is one of the main reasons for the importance of the
more advanced formulations of mechanics. They are a starting point for
the newer theories of relativity and quanta.
ELEMENTS OP NEWTONIAN MECHANICS
[CHAP. 1
1-2 Kinematics, the description of motion. Mechanics is the science
which studies the motions of physical bodies. We must first describe mo-
tions. Easiest to describe are the motions of a particle, that is, an object
whose size and internal structure are negligible for the problem with which
we are concerned. The earth, for example, could be regarded as a particle
for most problems in planetary motion, but certainly not for terrestrial
problems. We can describe the position of a particle by specifying a point
in space. This may be done by giving three coordinates. Usually, rec-
tangular coordinates are used. For a particle moving along a straight line
(Chapter 2) only one coordinate need be given. To describe the motion
of a particle, we specify the coordinates as functions of time :
one dimension : x(t),
three dimensions: x(t), y(t), z(t).
(1-D
The basic problem of classical mechanics is to find ways to determine func-
tions like these which specify the positions of objects as functions of time,
for any mechanical situation. The physical meaning of the function x(t)
is contained in the rules which tell us how to measure the coordinate re of a
particle at a time t. Assuming we know the meaning of x(t), or at least
that it has a meaning (this assumption, which we make in classical me-
chanics, is not quite correct according to quantum mechanics), we can
define the x-component of velocity v x at time t as*
_ _ dx
p Vx ~ x ~ ~dt
». = * = ^> (1-2)
~r
A
LA.
and, similarly,
dy
v,= z =
dz
dt'
three dimensions
h
one dimension
Fig. 1-1. Rectangular coordinates
specifying the position of a particle P
relative to an origin 0.
y-axis w e now d e fme the components of
acceleration a x , a y , a,,, as the deriva-
tives of the velocity components
with respect to time (we list several
equivalent notations which may be
used) :
dv x
dt
dVy
dt
dv z
~dJ
a x = v x = — 77 — x
a v = v v = -j? = y =
V, = -7T = Z
dt* '
d*y
dt*
dS
dt*'
(1-3)
* We shall denote a time derivative either by d/dt or by a dot. Both notations
are given in Eq. (1-2).
1-3] DYNAMICS. MASS AND FORCE 5
For many purposes some other system of coordinates may be more con-
venient for specifying the position of a particle. When other coordinate
systems are used, appropriate formulas for components of velocity and
acceleration must be worked out. Spherical, cylindrical, and plane polar
coordinates will be discussed in Chapter 3. For problems in two and three
dimensions, the concept of a vector is very useful as a means of represent-
ing positions, velocities, and accelerations. A systematic development of
vector algebra will be given in Section 3-1.
To describe a system of particles, we may specify the coordinates of
each particle in any convenient coordinate system. Or we may introduce
other kinds of coordinates, for example, the coordinates of the center of
mass, or the distance between two particles. If the particles form a rigid
body, the three coordinates of its center of mass and three angular coordi-
nates specifying its orientation in space are sufficient to specify its position.
To describe the motion of continuous matter, for example a fluid, we would
need to specify the density p(x, y, z, t) at any point (x, y, z) in space at each
instant t in time, and the velocity vector v(x, y, z, t) with which the matter
at the point (x, y, z) is moving at time t. Appropriate devices for describ-
ing the motion of physical systems will be introduced as needed.
1-3 Dynamics. Mass and force. Experience leads us to believe that
the motions of physical bodies are controlled by interactions between them
and their surroundings. Observations of the behavior of projectiles and
of objects sliding across smooth, well-lubricated surfaces suggest the idea
that changes in the velocity of a body are produced by interaction with its
surroundings. A body isolated from all interactions would have a con-
stant velocity. Hence, in formulating the laws of dynamics, we focus our
attention on accelerations.
Let us imagine two bodies interacting with each other and otherwise
isolated from interaction with their surroundings. As a rough approxima-
tion to this situation, imagine two boys, not necessarily of equal size, en-
gaged in a tug of war over a rigid pole on smooth ice. Although no two
actual bodies can ever be isolated completely from interactions with all
other bodies, this is the simplest kind of situation to think about and one
for which we expect the simplest mathematical laws. Careful experiments
with actual bodies lead us to conclusions as to what we should observe
if we could achieve ideal isolation of two bodies. We should observe that
the two bodies are always accelerated in opposite directions, and that the
ratio of their accelerations is constant for any particular pair of bodies no
matter how strongly they may be pushing or pulling each other. If we
measure the coordinates xi and x 2 of the two bodies along the line of their
accelerations, then
xi/x 2 = —ki2, (1^1)
6 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1
where k i2 is a positive constant characteristic of the two bodies concerned.
The negative sign expresses the fact that the accelerations are in opposite
directions.
Furthermore, we find that in general the larger or heavier or more mas-
sive body is accelerated the least. We find, in fact, that the ratio k i2 is
proportional to the ratio of the weight of body 2 to that of body 1. The
accelerations of two interacting bodies are inversely proportional to their
weights. This suggests the possibility of a dynamical definition of what
we shall call the masses of bodies in terms of their mutual accelerations.
We choose a standard body as a unit mass. The mass of any other body
is defined as the ratio of the acceleration of the unit mass to the accelera-
tion of the other body when the two are in interaction :
mi = k u = —xjxi, (1-5)
where m,- is the mass of body i, and body 1 is the standard unit mass.
In order that Eq. (1-5) may be a useful definition, the ratio fci 2 of the
mutual accelerations of two bodies must satisfy certain requirements. If
the mass defined by Eq. (1-5) is to be a measure of what we vaguely call
the amount of matter in a body, then the mass of a body should be the sum
of the masses of its parts, and this turns out to be the case to a very high
degree of precision. It is not essential, in order to be useful in scientific
theories, that physical concepts for which we give precise definitions should
correspond closely to any previously held common-sense ideas. However,
most precise physical concepts have originated from more or less vague
common-sense ideas, and mass is a good example. Later, in the theory
of relativity, the concept of mass is somewhat modified, and it is no longer
exactly true that the mass of a body is the sum of the masses of its parts.
One requirement which is certainly essential is that the concept of mass
be independent of the particular body which happens to be chosen as
having unit mass, in the sense that the ratio of two masses will be the
same no matter what unit of mass may be chosen. This will be true be-
cause of the following relation, which is found experimentally, between
the mutual acceleration ratios denned by Eq. (1-4) of any three bodies:
ki 2 k 23 k 31 = 1. (1-6)
Suppose that body 1 is the unit mass. Then if bodies 2 and 3 interact
with each other, we find, using Eqs. (1-4), (1-6), and (1-5),
x 2 /x 3 = —k 23
= - l/(*i2ft 81 ) (1-7)
= —k 13 /ki2
= — m 3 /m 2 .
1-4] newton's laws of motion 7
The final result contains no explicit reference to body 1, which was taken
to be the standard unit mass. Thus the ratio of the masses of any two
bodies is the negative inverse of the ratio of their mutual accelerations, in-
dependently of the unit of mass chosen.
By Eq. (1-7), we have, for two interacting bodies,
m 2 x 2 = — m x z x . (1-8)
This suggests that the quantity (mass X acceleration) will be important,
and we call this quantity the force acting on a body. The acceleration of
a body in space has three components, and the three components of force
acting on the body are
F x = mx, F y = my, F z = ml. (1-9)
The forces which act on a body are of various kinds, electric, magnetic,
gravitational, etc., and depend on the behavior of other bodies. In
general, forces due to several sources may act on a given body, and it is
found that the total force given by Eqs. (1-9) is the vector sum of the
forces which would be present if each source were present alone.
The theory of electromagnetism is concerned with the problem of de-
termining the electric and magnetic forces exerted by electrical charges
and currents upon one another. The theory of gravitation is concerned
with the problem of determining the gravitational forces exerted by
masses upon one another. The fundamental problem of mechanics is to
determine the motions of any mechanical system, given the forces acting
on the bodies which make up the system.
1-4 Newton's laws of motion. Isaac Newton was the first to give a
complete formulation of the laws of mechanics. Newton stated his famous
three laws as follows:*
(1) Every body continues in its state of rest or of uniform motion
in a straight line unless.it is compelled to change that state by forces
impressed upon it.
(2) Rate of change of momentum is proportional to the impressed
force, and is in the direction in which the force acts.
(3) To every action there is always opposed an equal reaction.
In the second law, momentum is to be defined as the product of the mass
and the velocity of the particle. Momentum, for which we use the symbol
* Isaac Newton, Mathematical Principles of Natural Philosophy and his System
of the World, tr. by F. Cajori (p. 13). Berkeley: University of California Press,
1934.
8 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1
p, has three components, denned along x-, y-, and z-axes by the equations
p x = mv x , p y = mv y , p z = mv z . (1-10)
The first two laws, together with the definition of momentum, Eqs. (1-10),
and the fact that the mass is constant by Eq. (1-4),* are equivalent to
Eqs. (1-9), which express them in mathematical form. The third law
states that when two bodies interact, the force exerted on body 1 by body 2
is equal and opposite in direction to that exerted on body 2 by body 1.
This law expresses the experimental fact given by Eq. (1-4), and can
easily be derived from Eq. (1-4) and from Eqs. (1-5) and (1-9).
The status of Newton's first two laws, or of Eqs. (1-9), is often the sub-
ject of dispute. We may regard Eqs. (1-9) as defining force in terms of
mass and acceleration. In this case, Newton's first two laws are not laws
at all but merely definitions of a new concept to be introduced in the theory.
The physical laws are then the laws of gravitation, electromagnetism, etc.,
which tell us what the forces are in any particular situation. Newton's
discovery was not that force equals mass times acceleration, for this is
merely a definition of "force." What Newton discovered was that the
laws of physics are most easily expressed in terms of the concept of force
defined in this way. Newton's third law is still a legitimate physical law
expressing the experimental result given by Eq. (1-4) in terms of the con-
cept of force. This point of view toward Newton's first two laws is con-
venient for many purposes and is often adopted. Its chief disadvantage is
that Eqs. (1-9) define only the total force acting on a body, whereas we
often wish to speak of the total force as a (vector) sum of component forces
of various kinds due to various sources. The whole science of statics,
which deals with the forces acting in structures at rest, would be unintelli-
gible if we took Eqs. (1-9) as our definition of force, for all accelerations are
zero in a structure at rest.
We may also take the laws of electromagnetism, gravitation, etc., to-
gether with the parallelogram law of addition, as defining "force. " Equa-
tions (1-9) then become a law connecting previously defined quantities.
This has the disadvantage that the definition of force changes whenever a
new kind of force (e.g., nuclear force) is discovered, or whenever modifica-
tions are made in electromagnetism or in gravitation. Probably the best
plan, the most flexible at least, is to take force as a primitive concept of
* In the theory of relativity, the mass of a body is not constant, but depends on
its velocity. In this case, law (2) and Eqs. (1-9) are not equivalent, and it turns
out that law (2) is the correct formulation. Force should then be equated to time
rate of change of momentum. The simple definition (1-5) of mass is not correct
according to the theory of relativity unless the particles being accelerated move
at low velocities.
1- 4] newton's laws of motion 9
our theory, perhaps defined operationally in terms of measurements with
a spring balance. Then Newton's laws are laws, and so are the laws of
theories of special forces like gravitation and electromagnetism.
Aside from the question of procedure in regard to the definition of force,
there are other difficulties in Newton's mechanics. The third law is not
always true. It fails to hold for electromagnetic forces, for example, when
the interacting bodies are far apart or rapidly accelerated and, in fact, it
fails for any forces which propagate from one body to another with finite
velocities. Fortunately, most of our development is based on the first
two laws. Whenever the third law is used, its use will be explicitly noted
and the results obtained will be valid only to the extent that the third law
holds.
Another difficulty is that the concepts of Newtonian mechanics are not
perfectly clear and precise, as indeed no concepts can probably ever be for
any theory, although we must develop the theory as if they were. An
outstanding example is the fact that no specification is made of the coordi-
nate system with respect to which the accelerations mentioned in the first
two laws are to be measured. Newton himself recognized this difficulty
but found no very satisfactory way of specifying the correct coordinate
system to use. Perhaps the best way to formulate these laws is to say
that there is a coordinate system with respect to which they hold, leaving
it to experiment to determine the correct coordinate system. It can be
shown that if these laws hold in any coordinate system, they hold also in
any coordinate system moving uniformly with respect to the first. This
is called the principle of Newtonian relativity, and will be proved in
Section 7-1, although the reader should find little difficulty in proving it
for himself.
Two assumptions which are made throughout classical physics are that
the behavior of measuring instruments is unaffected by their state of
motion so long as they are not rapidly accelerated, and that it is possible,
in principle at least, to devise instruments to measure any quantity with
as small an error as we please. These two assumptions fail in extreme
cases, the first at very high velocities, the second when very small magni-
tudes are to be measured. The failure of these assumptions forms the
basis of the theory of relativity and the theory of quantum mechanics,
respectively. However, for a very wide range of phenomena, Newton's
mechanics is correct to a very high degree of accuracy, and forms the
starting point at which the modern theories begin. Not only the laws but
also the concepts of classical physics must be modified according to the
modern theories. However, an understanding of the concepts of modern
physics is made easier by a clear understanding of the concepts of classical
physics. These difficulties are pointed out here in order that the reader
may be prepared to accept later modifications in the theory. This is not to
10
ELEMENTS OF NEWTONIAN MECHANICS
[CHAP. 1
say that Newton himself (or the reader either at this stage) ought to have
worried about these matters before setting up his laws of motion. Had
he done so, he probably never would have developed his theory at all. It
was necessary to make whatever assumptions seemed reasonable in order
to get started. Which assumptions needed to be altered, and when, and
in what way, could only be determined later by the successes and failures
of the theory in predicting experimental results.
1-5 Gravitation. Althoug
the motions of the planets
a property of physical bodies
to formulate a mathematical
Newton showed, by methods
planets could be quantitatively
every pair of bodies is
masses and inversely
them. In symbols,
there had been previous suggestions that
of falling bodies on earth might be due to
by which they attract one another, the first
theory of this phenomenon was Isaac Newton.
be considered later, that the motions of the
accounted for if he assumed that with
associated a force of attraction proportional to their
proportional to the square of the distance between
and
to]
F =
Gmim 2
(1-H)
where mi, m 2 are the masses of the attracting bodies, r is the distance be-
tween them, and G is a universal constant whose value according to ex-
periment is*
G = (6.670 ± 0.005) X 10
" 8 cm 3 -sec 2 -gm"
(1-12)
For a spherically symmetrical body, we shall show later (Section 6-2) that
the force can be computed as if all the mass were at the center. For a
small body of mass m at the surface of the earth, the force of gravitation
is therefore
F = mg, (1-13)
where
GM
9 =
fl2
= 980.2 cm-sec
—2
(1-14)
and M is the mass of the earth and R its radius. The quantity g has the
dimensions of an acceleration, and we can readily show by Eqs. (1-9) and
(1-13) that any freely falling body at the surface of the earth is accelerated
downward with an acceleration g.
The fact that the gravitational force on a body is proportional to its
mass, rather than to some other constant characterizing the body (e.g.,
its electric charge), is more or less accidental from the point of view of
Newton's theory. This fact is fundamental in the general theory of rela-
* Smithsonian Physical Tables, 9th ed., 1954.
1-6] UNITS AND DIMENSIONS 11
tivity. The proportionality between gravitational force and mass is proba-
bly the reason why the theory of gravitation is ordinarily considered a
branch of mechanics, while theories of other kinds of force are not.
Equation (1-13) gives us a more convenient practical way of measuring
mass than that contemplated in the original definition (1-5). We may
measure a mass by measuring the gravitational force on it, as in a spring
balance, or by comparing the gravitational force on it with that on a stand-
ard mass, as in the beam or platform balance; in other words, by weigh-
ing it.
1-6 Units and dimensions. In setting up a system of units in terms of
which to express physical measurements, we first choose arbitrary standard
units for a certain set of "fundamental" physical quantities (e.g., mass,
length, and time) and then define further derived units in terms of the
fundamental units (e.g., the unit of velocity is one unit length per unit of
time). It is customary to choose mass, length, and time as the funda-
mental quantities in mechanics, although there is nothing sacred in this
choice. We could equally well choose some other three quantities, or even
more or fewer than three quantities, as fundamental.
There are three systems of units in common use, the centimeter-gram-
second or cgs system, the meter-kilogram-second or mks system, and the
foot-pound-second or English system, the names corresponding to' the
names of the three fundamental units in each system.* Units for ether
kinds of physical quantities are obtained from their defining equations by
substituting the units for the fundamental quantities which occur. For
example, velocity, by Eq. (1-2),
dx
Vx = Tt'
is defined as a distance divided by a time. Hence the units of velocity are
cm/sec, m/sec, and ft/sec in the three above-mentioned systems, respec-
tively.
Similarly, the reader can show that the units of force in the three sys-
tems as given by Eqs. (1-9) are gm-cm-sec -2 , kgm-m-sec -2 , lb-ft-sec -2 .
These units happen to have the special names dyne, newton, and poundal,
respectively. Gravitational units of force are sometimes defined by re-
placing Eqs. (1-9) by the equations
F x = mx/g, F y = my/g, F z = mz/g, (1-15)
* In the mks system, there is a fourth fundamental unit, the coulomb of electri-
cal charge, which enters into the definitions of electrical units. Electrical units in
the cgs system are all defined in terms of centimeters, grams, and seconds. Elec-
trical units in the English system are practically never used.
(1-11)
12 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1
where g = 980.2 cm-sec -2 = 9.802 m-sec -2 = 32.16 ft-sec -2 is the stand-
ard acceleration of gravity at the earth's surface. Unit force is then that
force exerted by the standard gravitational field on unit mass. The names
gram-weight, kilogram-weight, pound-weight are given to the gravita-
tional units of force in the three systems. In the present text, we shall
write the fundamental law of mechanics in the form (1-9) rather than
(1-15) ; hence we shall be using the absolute units for force and not the
gravitational units.
Henceforth the question of units will rarely arise, since nearly all our
examples will be worked out in algebraic form. It is assumed that the
reader is sufficiently familiar with the units of measurement and their
manipulation to be able to work out numerical examples in any system of
units should the need arise.
In any physical equation, the dimensions or units of all additive terms
on both sides of the equation must agree when reduced to fundamental
units. As an example, we may check that the dimensions of the gravita-
tional constant in Eq. (1-11) are correctly given in the value quoted in
Eq. (1-12) :
„ = Gm 1 m 2
r 2
We substitute for each quantity the units in which it is expressed :
. _2 N (cm 3 -sec _2 -gm -1 )(gm)(gm) . _ 2x ,, 1C ^
(gm-cm-sec ) = - , „/ = (gm-cm-sec ). (1-16)
The check does not depend on which system of units we use so long as we
use absolute units of force, and we may check dimensions without any
reference to units, using symbols I, m, t for length, mass, time :
{mlr2) = (^- 2 m-*)(m)(m) = ( ^_ 2) (1 _ 1?)
When constant factors like G are introduced, we can, of course, always
make the dimensions agree in any particular equation by choosing appro-
priate dimensions for the constant. If the units in the terms of an equa-
tion do not agree, the equation is certainly wrong. If they do agree, this
does not guarantee that the equation is right. However, a check on dimen-
sions in a result will reveal most of the mistakes that result from algebraic
errors. The reader should form the habit of mentally checking the dimen-
sions of his formulas at every step in a derivation. When constants are
introduced in a problem, their dimensions should be worked out from the
first equation in which they appear, and used in checking subsequent steps.
1-7] SOME ELEMENTARY PROBLEMS IN MECHANICS 13
1-7 Some elementary problems in mechanics. Before beginning a sys-
tematic development of mechanics based on the laws introduced in this
chapter, we shall review a few problems from elementary mechanics in
order to fix these laws clearly in mind.
One of the simplest mechanical problems is that of finding the motion of
a body moving in a straight line, and acted upon by a constant force. If
the mass of the body is m and the force is F, we have, by Newton's second
law,
F = ma. (1-18)
The acceleration is then constant:
dv F
a = "^7 = — '
at m
(1-19)
If we multiply Eq. (1-19) by dt, we obtain an expression for the change in
velocity dv occurring during the short time dt:
F
dv = — dt. (1-20)
Integrating, we find the total change in velocity during the time t:
/ dv = / —dt, (1-21)
Jvq Jo m
v — v = — t, (1-22)
where v is the velocity at t = 0. If a; is the distance of the body from a
fixed origin, measured along its line of travel, then
We again multiply by dt and integrate to find x :
L dx =L( vo+ ^) dt > (i - 24)
xo ■'0
x = x + v t + \ — 1\ (1-25)
F,2
where x represents the position of the body at t = 0. We now have a
complete description of the motion. We can calculate from Eqs. (1-25)
and (1-22) the velocity of the body at any time t, and the distance it has
14
ELEMENTS OF NEWTONIAN MECHANICS
[CHAP. 1
traveled. A body falling freely near the surface of the earth is acted upon
by a constant force given by Eq. (1-13), and by no other force if air re-
sistance is negligible. In this case, if x is the height of the body above
some reference point, we have
F = —mg.
(1-26)
The negative sign appears because the force is downward and the positive
direction of x is upward. Substituting in Eqs. (1-19), (1-22), and (1-25),
we have the familiar equations
a =
v = v — gt,
x = x + v t — %gt 2 .
(1-27)
(1-28)
(1-29)
In applying Newton's law of motion, Eq. (1-18), it is essential to
decide first to what body the law is to be applied, then to insert the
mass m of that body and the total force F acting on it. Failure to keep
in mind this rather obvious point is the source of many difficulties, one of
which is illustrated by the horse-and-wagon dilemma. A horse pulls upon
a wagon, but according to Newton's third law the wagon pulls back with
an equal and opposite force upon the horse. How then can either the
wagon or the horse move? The reader who can solve Problem 4 at the
end of this chapter will have no difficulty answering this question.
Consider the motion of the system illustrated in Fig. 1-2. Two masses
mi and m 2 hang from the ends of a rope over a pulley, and we will suppose
that m 2 is greater than my. We take x as the distance from the pulley
mi
mig
ro2
mzg
Fig. 1-2. Atwood's machine.
1-7] SOME ELEMENTARY PROBLEMS IN MECHANICS 15
to m 2 . Since the length of the rope is constant, the coordinate x fixes
the positions of both m x and m 2 . Both move with the same velocity
. ■ = !• d- 3 °)
the velocity being positive when m^ is moving upward and m 2 is moving
downward. If we neglect friction and air resistance, the forces on mj
and m 2 are
F x = -vug + t, (1-31)
F 2 = m 2 g - t, (1-32)
where t is the tension in the rope. The forces are taken as positive when
they tend to produce a positive velocity dx/dt. Note that the terms involv-
ing t in these equations satisfy Newton's third law. The equations of
motion of the two masses are
— m x q + r = m x a, (1-33)
m 2 g — t = m 2 a, (1-34)
where a is the acceleration dv/dt, and is the same for both masses. By
adding Eqs. (1-33) and (1-34), we can eliminate r and solve for the accel-
eration :
d 2 x (m 2 — mi)
a = di? = (m, + m 2 ) 9- d-35)
The acceleration is constant and the velocity v and position x can be
found at any time t as in the preceding example. We can substitute for
a from Eq. (1-35) in either Eq. (1-33) or (1-34) and solve for the tension:
2m,\m 2
T = WH g - (1 ~ 36)
As a check, we note that if mx = m 2 , then a = and
t = mtf = m 2 g, (1-37)
as it should if the masses are in static equilibrium. As a matter of interest,
note that if m 2 » ntj, then
a = g, (1-38)
t = 2m ig . (1-39)
The reader should convince himself that these two results are to be ex-
pected in this case.
16
ELEMENTS OF NEWTONIAN MECHANICS
[CHAP. 1
\ (L
V\ mjf sin 6
mg cos 6/
^S>
V mg /
' Fig. 1-4. Resolution of forces into
Fig. 1-3. Forces acting on a brick components parallel and perpendicular
sliding down an incline. to the incline.
When several forces act on a body, its acceleration is determined by the
vector sum of the forces which act. Conversely, any force can be resolved
in any convenient manner into vector components whose vector sum is the
given force, and these components can be treated as separate forces acting
on the body.* As an example, we consider a brick of mass m sliding down
an incline, as shown in Fig. 1-3. The two forces which act on the brick are
the weight mg and the force F with which the plane acts on the brick.
These two forces are added according to the parallelogram law to give a
resultant R which acts on the brick:
R = ma.
(1-40)
Since the brick is accelerated in the direction of the resultant force, it is
evident that if the brick slides down the incline without jumping off or
penetrating into the inclined plane, the resultant force R must be directed
along the incline. In order to find R, we resolve each force into com-
ponents parallel and perpendicular to the incline, as in Fig. 1-4. The
force F exerted on the brick by the plane is resolved in Fig. 1-4 into two
components, a force N normal to the plane preventing the brick from
penetrating the plane, and a force / parallel to the plane, and opposed to
* A systematic development of vector algebra will be given in Chapter 3. Only
an understanding of the parallelogram law for vector addition is needed for the
present discussion.
1-7] SOME ELEMENTARY PROBLEMS IN MECHANICS 17
the motion of the brick, arising from the friction between the brick and the
plane. Adding parallel components, we obtain
R = mg sin 8 — f, (1^1)
and
= N — mg cos 0. (1-42)
If the frictional force / is proportional to the normal force N, as is often
approximately true for dry sliding surfaces, then
/ = fiN = fjimg cos d, iX~4S)
where p, is the coefficient of friction. Using Eqs. (1-43), (1-il), and (1-40),
we can calculate the acceleration :
a = g (sin — n cos 0). (1-44)
The velocity and position can now be found as functions of the time t,
as in the first example. Equation (1-44) holds only when the brick is
sliding down the incline. If it is sliding up the incline, the force / will
oppose the motion, and the second term in Eq. (1-44) will be positive.
This could only happen if the brick were given an initial velocity up the
incline. If the brick is at rest, the frictional force / may have any value
up to a maximum fi,N:
f < H*N, (1-45)
where /*„, the coefficient of static friction, is usually greater than p. In
this case R is zero, and
/ = mg sin 6 < n s mg cos 0. (1-46)
According to Eq. (1-46), the angle of the incline must not be greater than
a limiting value 6 r , the angle of repose:
tan < tan 6 r = ju s . (1-47)
If is greater than 9 T , the brick cannot remain at rest.
If a body moves with constant speed v around a circle of radius r, its
acceleration is toward the center of the circle, as we shall prove in Chapter
3, and is of magnitude
v 2
« = -• (1-48)
Such a body must be acted on by a constant force toward the center.
This centripetal force is given by
F = ma = ?y- ■ (1-49)
18 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1
Note that mv 2 /r is not a "centrifugal force" directed away from the center,
but is mass times acceleration and is directed toward the center, as is the
centripetal force F. As an example, the moon's orbit around the earth is
nearly circular, and if we assume that the earth is at rest at the center, then,
by Eq. (1-11), the force on the moon is
F = «M2 , (1-50)
where M is the mass of the earth and m that of the moon. We can ex-
press this force in terms of the radius R of the earth and the acceleration
g of gravity at the earth's surface by substituting for GM from Eq. (1-14) :
F= M^!. (1-51)
The speed v of the moon is
v = 2 ^, d-52)
where T is the period of revolution. Substituting Eqs. (1-51) and (1-52)
in Eq. (1-49), we can find r:
<* = *%£■ («8>
This equation was first worked out by Isaac Newton in order to check his
inverse square law of gravitation.* It will not be quite accurate because
the moon's orbit is not quite circular, and also because the earth does not
remain at rest at the center of the moon's orbit, but instead wobbles
slightly due to the attraction of the moon. By Newton's third law, this
attractive force is also given by Eq. (1-51). Since the earth is much
heavier than the moon, its acceleration is much smaller, and Eq. (1-53)
will not be far wrong. The exact treatment of this problem is given in
Section 4-7. Another small error is introduced by the fact that g, as de-
termined experimentally, includes a small effect due to the earth's rota-
tion. (See Section 7-3.) If we insert the measured values,
g = 980.2 cm-sec -2 ,
R = 6,368 kilometers,
T = 27J days,
we obtain, from Eq. (1-53),
r = 383,000 kilometers.
! Isaac Newton, op. cit., p. 407.
PROBLEMS 19
The mean distance to the moon according to modern measurements is
r = 385,000 kilometers.
The values of r and R available to Newton would not have given such close
agreement.
Problems
1. Compute the gravitational force of attraction between an electron and a
proton at a separation of 0.5 A (1 A = 10 -8 cm). Compare with the electrostatic
force of attraction at the same distance.
2. The coefficient of viscosity ij is defined by the equation
F dv
where F is the frictional force acting across an area A in a moving fluid, and dv is
the difference in velocity parallel to A between two layers of fluid a distance ds
apart, ds being measured perpendicular to A. Find the units in which the vis-
cosity ij would be expressed in the foot-pound-second, cgs, and mks systems. Find
the three conversion factors for converting coefficients of viscosity from one of
these systems to another.
3. A motorist is approaching a green traffic light with speed »o, when the
light turns to amber, (a) If his reaction time is r, during which he makes his
decision to stop and applies his foot to the brake, and if his maximum braking
deceleration is o, what is the minimum distance s mm from the intersection at the
moment the light turns to amber in which he can bring his car to a stop? (b) If
the amber light remains on for a time t before turning red, what is the maximum
distance Smax from the intersection at the moment the light turns to amber such
that he can continue into the intersection at speed vo without running the red
light? (c) Show that if his initial speed vq is greater than
V0 max = 2a(< — t),
there will be a range of distances from the intersection such that he can neither
stop in time nor continue through without running the red light, (d) Make some
reasonable estimates of t, t, and o, and calculate vo max in miles per hour. If vo =
§!>o max, calculate s mm and s ma x.
4. A boy of mass m pulls (horizontally) a sled of mass M . The coefficient of
friction between sled and snow is /x. (a) Draw a diagram showing all forces acting
on the boy and on the sled, (b) Find the horizontal and vertical components oA
each force at a moment when boy and sled each have an acceleration a. (c) If
the coefficient of static friction between the boy's feet and the ground is ju„ what
is the maximum acceleration he can give to himself and the sled, assuming trac-
tion to be the limiting factor?
20 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1
5. A floor mop of mass m is pushed with a force F directed along the handle,
which makes an angle 6 with the vertical. The coefficient of friction with the floor
is fi. (a) Draw a diagram showing all forces acting on the mop. (b) For given
6, n, find the force F required to slide the mop with uniform velocity across the
floor, (c) Show that if is less than the angle of repose, the mop cannot be started
across the floor by pushing along the handle. Neglect the mass of the mop handle.
6. A box of mass m slides across a horizontal table with coefficient of friction
fi. The box is connected by a rope which passes over a pulley to a body of mass M
hanging alongside the table. Find the acceleration of the system and the tension
in the rope.
7. The brick shown in Figs. 1-3 and 1-4 is given an initial velocity t>o up the
incline. The angle 6 is greater than the angle of repose. Find the distance the
brick moves up the incline, and the time required for it to slide up and back to its
original position.
8. A curve in a highway of radius of curvature r is banked at an angle 6 with
the horizontal. If the coefficient of friction is /x 8 , what is the maximum speed with
which a car can round the curve without skidding?
9. Assuming the earth moves in a circle of radius 93,000,000 miles, with a
period of revolution of one year, find the mass of the sun in tons.
10. (a) Compute the mass of the earth from its radius and the values of g
and G. (b) Look up the masses and distances of the sun and moon and compute
the force of attraction between earth and sun and between earth and moon.
Check your results by making a rough estimate of the ratio of these two forces
from a consideration of the fact that the former causes the earth to revolve about
the sun once a year, whereas the latter causes the earth to wobble in a small
circle, approximately once a month, about the common center of gravity of the
earth-moon system.
11. The sun is about 25,000 light years from the center of the galaxy, and
travels approximately in a circle at a speed of 175 mi/sec. Find the approximate
mass of the galaxy by assuming that the gravitational force on the sun can be
calculated as if all the mass of the galaxy were at its center. Express the result as
a ratio of the galactic mass to the sun's mass. (You do not need to look up either
G or the sun's mass to do this problem if you compare the revolution of the sun
around the galactic center with the revolution of the earth about the sun.)
CHAPTER 2
MOTION OF A PARTICLE IN ONE DIMENSION
2-1 Momentum and energy theorems. In this chapter, we study the
motion of a particle of mass m along a straight line, which we will take to
be the x-axis, under the action of a force F directed along the z-axis. The
discussion will be applicable, as we shall see, to other cases where the
motion of a mechanical system depends on only one coordinate, or where
all but one coordinate can be eliminated from the problem.
The motion of the particle is governed, according to Eqs. (1-9), by the
equation
m§=F. (2-1)
Before considering the solution of Eq. (2-1), we shall define some concepts
which are useful in discussing mechanical problems and prove some
simple general theorems about one-dimensional motion. The linear mo-
mentum p, according to Eq. (1-10), is denned as
p = mo = m -37 • (2-2)
From Eq. (2-1), using Eq. (2-2) and the fact that m is constant, we obtain
dp w
dt
(2-3)
This equation states that the time rate of change of momentum is equal
to the applied force, and is, of course, just Newton's second law. We may
call it the (differential) momentum theorem. If we multiply Eq. (2-3)
by dt and integrate from ti to t 2 , we obtain an integrated form of the
momentum theorem:
P2 — Pi
[ H Fdt. (2-4)
Equation (2-4) gives the change in momentum due to the action of the
force F between the times t t and t 2 - The integral on the right is called
the impulse delivered by the force F during this time; F must be known
as a function of t alone in order to evaluate the integral. If F is given
as F(x, v, f), then the impulse can be computed for any particular given
motion x(t), v(t).
21
22 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
A quantity which will turn out to be of considerable importance is the
kinetic energy, denned (in classical mechanics) by the equation
(2-5)
T
= imv 2 .
If
we multiply Eq. (2-1) by v,
we
mv
obtain
dv „
dt = FV >
or
s»
mv 2
. dT
' dt
Fv
(2-6)
Equation (2-6) gives the rate of change of kinetic energy, and may be
called the (differential) energy theorem. If we multiply by dt and inte-
grate from <i to t 2 , we obtain the integrated form of the energy theorem:
rl 2
T x = f 2 Fv dt. (2-7)
Jti
Equation (2-7) gives the change in energy due to the action of the force F
between the times ti and t%. The integral on the right is called the work
done by the force during this time. The integrand Fv on the right is the
time rate of doing work, and is called the power supplied by the force F.
In general, when F is given as F(x, v, t), the work can only be computed
for a particular specified motion x(t), v(t). Since v = dx/dt, we can re-
write the work integral in a form which is convenient when F is known
as a function of x:
T 2 - T x = f F dx. (2-8)
Jxi
2-2 Discussion of the general problem of one-dimensional motion. If
the force F is known, the equation of motion (2-1) becomes a second-order
ordinary differential equation for the unknown function x(t). The force F
may be known as a function of any or all of the variables t, x, and v. For
any given motion of a dynamical system, all dynamical variables (x, v, F,
p, T, etc.) associated with the system are, of course, functions of the
time t, that is, each has a definite value at any particular time t. However,
in many cases a dynamical variable such as the force may be known to
bear a certain functional relationship to x, or to v, or to any combination
of x, v, and t. As an example, the gravitational force acting on a body
falling from a great height above the earth is known as a function of the
height above the earth. The frictional drag on such a body would depend
on its speed and on the density of the air and hence on the height above
the earth; if atmospheric conditions are changing, it would also depend
on t. If F is given as F(x, v, t), then when x(t) and v(t) are known, these
2-2] THE GENERAL PROBLEM 23
functions can be substituted to give F as a function of the time t alone;
however, in general, this cannot be done until after Eq. (2-1) has been
solved, and even then the function F(t) may be different for different
possible motions of the particle. In any case, if F is given as F(x, v, t)
(where F may depend on any or all of these variables), then Eq. (2-1)
becomes a definite differential equation to be solved:
<^ = ±F(x,x,t). (2-9)
This is the most general type of second-order ordinary differential equa-
tion, and we shall be concerned in this- chapter with studying its solutions
and their applications to mechanical problems.
Equation (2-9) applies to all possible motions of the particle under the
action of the specified force. In general, there will be many such motions,
for Eq. (2-9) prescribes only the acceleration of the particle at every in-
stant in terms of its position and velocity at that instant. If we know
the position and velocity of a particle at a certain time, we can determine
its position a short time later (or earlier). Knowing also its acceleration,
we can find its velocity a short time later. Equation (2-9) then gives the
acceleration a short time later. In this manner, we can trace out the past
or subsequent positions and velocities of a particle if its position x and
velocity v are known at any one time t . Any pair of values of x and v
will lead to a possible motion of the particle. We call t Q the initial instant,
although it may be any moment in the history of the particle, and the
values of x and v at t we call the initial conditions. Instead of specifying
initial values for x and v, we could specify initial values of any two quan-
tities from which x and v can be determined; for example, we may specify
x and the initial momentum p — mv . These initial conditions, together
with Eq. (2-9), then represent a perfectly definite problem whose solu-
tion should be a unique function x(t) representing the motion of the
particle under the specified conditions.
The mathematical theory of second-order ordinary differential equa-
tions leads to results in agreement with what we expect from the nature
of the physical problem in which the equation arises. The theory asserts
that, ordinarily, an equation of the form (2-9) has a unique continuous
solution x(t) which takes on given values x Q and v of x and x at any chosen
initial value t of t. "Ordinarily" here means, as far as the beginning
mechanics student is concerned, "in all cases of physical interest."* The
properties of differential equations like Eq. (2-9) are derived in most
* For a rigorous mathematical statement of the conditions for the existence of
a solution of Eq. (2-9), see W. Leighton, An Introduction to the Theory of Differen-
tial Equations. New York: McGraw-Hill, 1952. (Appendix 1.)
24 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
treatises on differential equations. We know that any physical problem
must always have a unique solution, and therefore any force function
F(x, x, t) which can occur in a physical problem will necessarily satisfy the
required conditions for those values of x, x, t of physical interest. Thus
ordinarily we do not need to worry about whether a solution exists. How-
ever, most mechanical problems involve some simplification of the actual
physical situation, and it is possible to oversimplify or otherwise distort
a physical problem in such a way that the resulting mathematical problem
no longer possesses a unique solution. The general practice of physicists
in mechanics and elsewhere is to proceed, ignoring questions of mathe-
matical rigor. On those fortunately rare occasions when we run into diffi-
culty, we then consult our physical intuition, or check, our lapses of rigor,
until the source of the difficulty is discovered. Such a procedure may
bring shudders to the mathematician, but it is the most convenient and
rapid way to apply mathematics to the solution of physical problems.
The physicist, while he may proceed in a nonrigorous fashion, should never-
theless be acquainted with the rigorous treatment of the mathematical
methods which he uses.
The existence theorem for Eq. (2-9) guarantees that there is a unique
mathematical solution to this equation for all cases which will arise in
practice. In some cases the exact solution can be found by elementary
methods. Most of the problems considered in this text will be of this
nature. Fortunately, many of the most important mechanical problems
in physics can be solved without too much difficulty. In fact, one of the
reasons why certain problems are considered important is that they can
be easily solved. The physicist is concerned with discovering and verify-
ing the laws of physics. In checking these laws experimentally, he is free,
to a large extent, to choose those cases where the mathematical analysis
is not too difficult to carry out. The engineer is not so fortunate, since
his problems are selected not because they are easy to solve, but because
they are of practical importance. In engineering, and often also in physics,
many cases arise where the exact solution of Eq. (2-9) is difficult or im-
possible to obtain. In such cases various methods are available for obtain-
ing at least approximate answers. The reader is referred to courses and
texts on differential equations for a discussion of such methods.* From
the point of view of theoretical mechanics, the important point is that
a solution always does exist and can be found, as accurately as desired.
We shall restrict our attention to examples which can be treated by
simple methods.
* W. E. Milne, Numerical Calculus. Princeton: Princeton University Press,
1949. (Chapter 5.)
H. Levy and E. A. Baggott, Numerical Solutions of Differential Equations.
New York: Dover Publications, 1950.
2-3] APPLIED FORCE DEPENDING ON THE TIME 25
2-3 Applied force depending on the time. If the force F is given as a
function of the time, then the equation of motion (2-9) can be solved in
the following manner. Multiplying Eq. (2-9) by dt and integrating from
an initial instant t to any later (or earlier) instant t, we obtain Eq. (2-4),
which in this case we write in the form
mv ^- mv
/" F(t)dt. (2-10)
J to
Since F(t) is a known function of t, the integral on the right can, at least
in principle, be evaluated and the right member is then a function of t
(and t ). We solve for v :
dr 1 /"'
V = -dl = V o + m F ® dL ( 2 - U )
J to
Now multiply by dt and integrate again from t to t:
x — x = v (t — t ) +— I \ J F(t) dt
m Lo L/io
dt. (2-12)
To avoid confusion, we may rewrite the variable of integration as t' in
the first integral and t" in the second:
x = xo + v (t - to) + — l dt" I F(t') dt'. (2-13)
m I to J tB
This gives the required solution x(t) in terms of two integrals which can
be evaluated when F(t) is given. A definite integral can always be evalu-
ated. If an explicit formula for the integral cannot be found, then at
least it can always be computed as accurately as we please by numerical
methods. For this reason, in the discussion of a general type of problem
such as the one above, we ordinarily consider the problem solved when
the solution has been expressed in terms of one or more definite integrals.
In a practical problem, the integrals would have to be evaluated to obtain
the final solution in usable form.*
* The reader who has studied differential equations may be disturbed by the
appearance of three constants, to, »o, and xo, in the solution (2-13), whereas the
general solution of a second-order differential equation should contain only two
arbitrary constants. Mathematically, there are only two independent constants
in Eq. (2-13), an additive constant containing the terms xo — voto plus a term
from the lower limit of the last integral, and a constant multiplying t containing
the term wo plus a term from the lower limit of the first integral. Physically, we
can take any initial instant to, and then just two parameters xo and t»o are re-
quired to specify one out of all possible motions subject to the given force.
26 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
Problems in which F is given as a function of t usually arise when we
seek to find the behavior of a mechanical system under the action of some
external influence. As an example, we consider the motion of a free elec-
tron of charge — e when subject to an oscillating electric field along the
ic-axis:
E x = E cos (ut + 0). (2-14)
The force on the electron is
F = —eE x = —eE cos (ut + 0). (2-15)
The equation of motion is
m ^ = -eE cos (ut + 0). (2-16)
We multiply by dt and integrate, taking t = :
dx . eE sin eEo .,.,„■> ,„ ,«
v = -jt = v H y sin (ut + 0). (2-17)
dt mu mu
Integrating again, we obtain
eEn cos . ( . eE sin \ , . eE . . ,„ ,.,
x = x — u „ h V ^o H 1 1 ^ ^ cos (at + 6). (2-18)
If the electron is initially at rest at x = 0, this becomes
a; = ^— 5 1 -\ ^ cos (ut + 0). (2-19)
mw 2 mu mu 2
It is left to the reader to explain physically the origin of the constant term
and the term linear in t in Eq. (2-19) in terms of the phase of the electric
field at the initial instant. How do the terms in Eq. (2-19) depend on
e, m, E , and w? Explain physically. Why does the oscillatory term turn
out to be out of phase with the applied force?
The problem considered here is of interest in connection with the propa-
gation of radio waves through the ionosphere, which contains a high density
of free electrons. Associated with a radio wave of angular frequency w is
an electric field which may be given by Eq. (2-14). The oscillating term in
Eq. (2-18) has the same frequency w and is independent of the initial
conditions. This coherent oscillation of the free electrons modifies the
propagation of the wave. The nonoscillating terms in Eq. (2-18) depend
on the initial conditions, and hence on the detailed motion of each electron
as the wave arrives. These terms cannot contribute to the propagation
characteristics of the wave, since they do not oscillate with the frequency
of the wave, although they may affect the leading edge of the wave which
2-3] APPLIED FOBCE DEPENDING ON THE TIME 27
arrives first. We see that the oscillatory part of the displacement x is 180°
out of phase with the applied force due to the electric field. Since the elec-
tron has a negative charge, the resulting electric polarization is 180° out
of phase with the electric field. The result is that the dielectric constant
of the ionosphere is less than one. (In an ordinary dielectric at low fre-
quencies, the charges are displaced in the direction of the electric force
on them, and the dielectric constant is greater than one.) Since the veloc-
ity of light is
v = c(ne)- 112 , (2-20)
where c = 3 X 10 10 cm/sec and e and /x are the dielectric constant and
magnetic permeability respectively, and since /x = 1 here, the (phase)
velocity v of radio waves in the ionosphere is greater than the velocity c
of electromagnetic waves in empty space. Thus waves entering the
ionosphere at an angle are bent back toward the earth. The effect is seen
to be inversely proportional to w 2 , so that for high enough frequencies,
the waves do not return to the earth but pass out through the ionosphere.
Only a slight knowledge of electromagnetic theory is required to carry this dis-
cussion through mathematically.* The dipole moment of the electron displaced
from its equilibrium position is
e 2 2
—ex = = E cos (tot + 0) = e —= E x (2-21)
(wo 2 mw 2
if we consider only the oscillating term. If there are N electrons per cm 3 , the
total dipole moment per unit volume is
Ne 2
P x = 5 E x . (2-22)
The electric displacement is
D x = E x + 1tP x = (l - ^r) E *- (2-23)
Since the dielectric constant is denned by
D x = eE x , (2-24)
we conclude that
and since /i = 1,
....(l-M^)" 1 ". (2-26)
* See, e.g., G. P. Harnwell, Principles of Electricity and Electromagnetism, 2nd
ed. New York: McGraw-Hill, 1949. (Section 2.4.)
28 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
2-4 Damping force depending on the velocity. Another type of force
which allows an easy solution of Eq. (2-9) is the case when F is a function
of v alone :
m ~ = F(v). (2-27)
To solve, we multiply by [m^(y)] -1 dt and integrate from t to t:
dv t — to
f
F(v)
(2-28)
The integral on the left can be evaluated, in principle at least, when F(v)
is given, and an equation containing the unknown v results. If this equa-
tion is solved for v (we assume in general discussions that this can always
be done), we will have an equation of the form
The solution for x is then
x = x +
C«( v -hr) dL (2 " 30)
In the case of one-dimensional motion, the only important kinds of forces
which depend on the velocity are frictional forces. The force of sliding or
rolling friction between dry solid surfaces is nearly constant for a given
pair of surfaces with a given normal force between them, and depends on
the velocity only in that its direction is always opposed to the velocity.
The force of friction between lubricated surfaces or between a solid body
and a liquid or gaseous medium depends on the velocity in a complicated
way, and the function F(v) can usually be given only in the form of a
tabulated summary of experimental data. In certain cases and over
certain ranges of velocity, the frictional force is proportional to some fixed
power of the velocity:
F = (=F)bv n . (2-31)
If n is an odd integer, the negative sign should be chosen in the above
equation. Otherwise the sign must be chosen so that the force has the
opposite sign to the velocity v . The frictional force is always opposed to
the velocity, and therefore does negative work, i.e., absorbs energy from
the moving body. A velocity-dependent force in the same direction as
the velocity would represent a source of energy; such cases do not often
occur.
2-4] DAMPING FORCE DEPENDING ON THE VELOCITY 29
As an example, we consider the problem of a boat traveling with initial
velocity v , which shuts off its engines at t = when it is at the position
xo = 0. We assume the force of friction given by Eq. (2-31) with n = 1 :
dv
m-TT=—bv. (2-32)
We solve Eq. (2-32), following the steps outlined above [Eqs. (2-27)
through (2-30)]:
rv
Jvo V
It,
m
=
1.
v e~
-btjm
dt
=
mvQ
b
(1
— e"
—bt/m
, v b
In — = t,
Vq m
v = v e- btlm . (2-33)
We see that as t — ► oo , v — * 0, as it should, but that the boat never comes
completely to rest in any finite time. The solution for x is
(2-34)
As t — * oo , x approaches the limiting value
*, = ^. (2-35)
Thus we can specify a definite distance that the boat travels in stopping.
Although according to the above result, Eq. (2-33), the velocity never
becomes exactly zero, when t is sufficiently large the velocity becomes so
small that the boat is practically stopped. Let us choose some small
velocity v 3 such that when v < v„ we are willing to regard the boat as
stopped (say, for example, the average random speed given to an anchored
boat by the waves passing by it). Then we can define the time t a required
for the boat to stop by
v s = v e- bt '' m , < s = ^ln^- (2-36)
Vg
Since the logarithm is a slowly changing function, the stopping time t s will
not depend to any great extent on precisely what value of v e we choose so
long as it is much smaller than v . It is often instructive to expand solu-
tions in a Taylor series in t. If we expand the right side of Eqs. (2-33)
30 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
and (2-34) in power series in t, we obtain*
v = v -^t + ---, (2-37)
x = v t-± b -^t 2 + ---. (2-38)
Note that the first two terms in the series for v and x are just the formulas
for a particle acted on by a constant force —too, which is the initial value
of the frictional force in Eq. (2-32). This is to be expected, and affords a
fairly good check on the algebra which led to the solution (2-34). Series
expansions are a very useful means of obtaining simple approximate for-
mulas valid for a short range of time.
The characteristics of the motion of a body under the action of a fric-
tional force as given by Eq. (2-31) depend on the exponent n. In general,
a large exponent n will result in rapid initial slowing but slow final stopping,
and vice versa, as one can see by sketching graphs of F vs. v for various
values of n. For small enough values of n, the velocity comes to zero in
a finite time. For large values of n, the body not only requires an infinite
time, but travels an infinite distance before stopping. This disagrees
with ordinary experience, an indication that while the exponent n may be
large at high velocities, it must become smaller at low velocities. The
exponent n = 1 is often assumed in problems involving friction, particu-
larly when friction is only a small effect to be taken into account approxi-
mately. The reason for taking n = 1 is that this gives easy equations
to solve, and is often a fairly good approximation when the frictional
force is small, provided b is properly chosen. -— '
2-5 Conservative force depending on position. Potential energy. One
of the most important types of motion occurs when the force F is a func-
* The reader who has not already done so should memorize the Taylor series
for a few simple functions like
2 3 4
l + x + X - + X ' X
2 ' 2-3 ' 2-3-4
ln(l + x) = *_|- + ?L_fL +
(i + xr = i + ^ + ^^/ + w(w - 1)( r- 2) x 3 + ---.
These three series are extremely useful in obtaining approximations to compli-
cated formulas, valid when x is small.
2-5] CONSERVATIVE FORCE DEPENDING ON POSITION 31
tion of the coordinate x alone :
dv
m Tt = F(x). (2-39)
We have then, by the energy theorem (2-8),
imv 2 — \rrw\ = f" F(x) dx. (2-40)
Jx
The integral on the right is the work done by the force when the particle
goes from x to x. We now define the potential energy V(x) as the work
done by the force when the particle goes from x to some chosen standard
point x s :
V(x) = f ° F(x) dx= - f F(x) dx. (2-41)
Jx Jx,
The reason for calling this quantity potential energy will appear shortly.
In terms of V(x), we can write the integral in Eq. (2-40) as follows:
f F(x) dx = -V(x) + V(x ). (2-42)
Jx
With the help of Eq. (2-42), Eq. (2-40) can be written
\mv 2 + V{x) = \mvl + V(x ). (2^3)
The quantity on the right depends only on the initial conditions and is
therefore constant during the motion. It is called the total energy E, and
we have the law of conservation of kinetic plus potential energy, which
holds, as we can see, only when the force is a function of position alone :
\mo 2 + V(x) = T + V = E. (2-44)
Solving for v, we obtain
v =tt->M [E ~ nx)]U2 - <*-«)
The function x(t) is to be found by solving for x the equation
\?i..
[E - V(x)]- 112 dx = t- t . (2-46)
In this case, the initial conditions are expressed in terms of the constants
E and xq.
In applying Eq. (2-46), and in taking the indicated square root in the
integrand, care must be taken to use the proper sign, depending on whether
32 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
the velocity v given by Eq. (2-45) is positive or negative. In cases where
v is positive during some parts of the motion and negative during other
parts, it may be necessary to carry out the integration in Eq. (2-46)
separately for each part of the motion.
From the definition (2-41) we can express the force in terms of the
potential energy:
This equation can be taken as expressing the physical meaning of the po-
tential energy. The potential energy is a function whose negative
derivative gives the force. The effect of changing the coordinate of the
standard point x s is to add a constant to V(x). Since it is the derivative
of V which enters into the dynamical equations as the force, the choice
of standard point x, is immaterial. A constant can always be added to
the potential V(x) without affecting the physical results. (The same
constant must, of course, be added to E.)
As an example, we consider the problem of a particle subject to a
linear restoring force, for example, a mass fastened to a spring:
F = -kx. (2-48)
The potential energy, if we take x s — 0, is
rx
V{x) = — I —kx dx
Jo
= \kx 2 . (2-49)
Equation (2-46) becomes, for this case, with t = 0,
(E - %kx 2 )~ U2 dx = t. (2-50)
\?ix„
'10
Now make the substitutions
sin d = x yjzjj ' (2-51)
W = JA, (2-52)
so that
Jf f (E - %kx 2 )- 112 dx = i f de = Ue-e ),
\ ^ Jxq w Je «
>x
and, by Eq. (2-50),
6 = <at + O .
2-5] CONSERVATIVE FORCE DEPENDING ON POSITION 33
We can now solve for x in Eq. (2-51) :
x = J-r- sin = A sin (wi + O ), (2-53)
where
F- ( 2 ~ 54 )
Thus the coordinate x oscillates harmonically in time, with amplitude A
and frequency co/27r. The initial conditions are here determined by the
constants A and O , which are related to E and x by
E = %JcA 2 , (2-55)
x Q = A sin O . (2-56)
Notice that in this example we meet the sign difficulty in taking the
square root in Eq. (2-50) by replacing (1 — sin 2 0) _1/2 by (cos0) _1 , a
quantity which can be made either positive or negative as required by
choosing in the proper quadrant.
A function of the dependent variable and its first derivative which
is constant for all solutions of a second-order differential equation, is called
a first integral of the equation. The function %mx 2 + V(x) is called the
energy integral of Eq. (2-39). An integral of the equations of motion of
a mechanical system is also called a constant of the motion. In general,
any mechanical problem can be solved if we can find enough first inte-
grals, or constants of the motion.
Even in cases where the integral in Eq. (2-16) cannot easily be evalu-
ated or the resulting equation solved to give an explicit solution for x(t),
the energy integral, Eq. (2-44), gives us useful information about the
solution. For a given energy E, we see from Eq. (2-15) that the particle
is confined to those regions on the x-axis where V(x) < E. Furthermore,
the velocity is proportional to the square root of the difference between
E and V(x). Hence, if we plot V(x) versus x, we can give a good qualita-
tive description of the kinds of motion that are possible. For the potential-
energy function shown in Fig. 2-1 we note that the least energy possible
is E . At this energy, the particle can only be at rest at x Q . With a
slightly higher energy E u the particle can move between xi and x 2 ; its
velocity decreases as it approaches xi or x 2 , and it stops and reverses its
direction when it reaches either Xi or x 2 , which are called turning points
of the motion. With energy E 2 , the particle may oscillate between turn-
ing points x 3 and x iy or remain at rest at x 5 . With energy E 3 , there are
four turning points and the particle may oscillate in either of the two
34
MOTION OF A PARTICLE IN ONE DIMENSION
[CHAP. 2
xa xzx\ x« X2 Xi Xi
Fig. 2-1. A potential-energy function for one-dimensional motion.
potential valleys. With energy E t , there is only one turning point; if
the particle is initially traveling to the left, it will turn at x 6 and return
to the right, speeding up over the valleys at x and x 5 , and slowing down
over the hill between. At energies above E 5 , there are no turning points
and the particle will move in one direction only, varying its speed accord-
ing to the depth of the potential at each point.
A point where V(x) has a minimum is called a point of stable equilibrium.
A particle at rest at such a point will remain at rest. If displaced a slight
distance, it will experience a restoring force tending to return it, and it
will oscillate about the equilibrium point. A point where V(x) has a maxi-
mum is called a point of unstable equilibrium. In theory, a particle at
rest there can remain at rest, since the force is zero, but if it is displaced
the slightest distance, the force acting on it will push it farther away from
the unstable equilibrium position. A region where V(x) is constant is
called a region of neutral equilibrium, since a particle can be displaced
slightly without suffering either a restoring or a repelling force.
This kind of qualitative discussion, based on the energy integral, is
simple and very useful. Study this example until you understand it well
enough to be able to see at a glance, for any potential energy curve, the
types of motion that are possible.
It may be that only part of the force on a particle is derivable from a
potential function V(x). Let F' be the remainder of the force:
F =
dV
dx
+ F'.
(2-57)
In this case the energy (T -\- V) is no longer constant. If we substitute F
2-6] FALLING BODIES 35
from Eq. (2-57) in Eq. (2-1), and multiply by dx/dt, we have, after
rearranging terms,
I (T + V) = F'v. (2-58)
The time rate of change of kinetic plus potential energy is equal to the
power delivered by the additional force F'.
2-6 Falling bodies. One of the simplest and most commonly occurring
types of one-dimensional motion is that of falling bodies. We take up
this type of motion here as an illustration of the principles discussed in
the preceding sections.
A body falling near the surface of the earth, if we neglect air resistance,
is subject to a constant force
F = —mg, (2-59)
where we have taken the positive direction as upward. The equation of
motion is
d 2 x
™- d ~p = — mg. (2-60)
The solution may be obtained by any of the three methods discussed
in Sections 2-3, 2-4, and 2-5, since a constant force may be considered
as a function of either t, v, or x. The reader will find it instructive to
solve the problem by all three methods. We have already obtained the
result in Chapter 1 [Eqs. (1-28) and (1-29)].
In order to include the effect of air resistance, we may assume a fric-
tional force proportional to v, so that the total force is
F = — mg — bv. (2-61)
The constant & will depend on the size and shape of the falling body,
as well as on the viscosity of the air. The problem must now be treated
as a case of F(v):
dv ,
m -^ = — mg — bv. (2-62)
Taking v = at t = 0, we proceed as in Section 2-4 [Eq. (2-28)]:
dv bt
f
Jo
! v + (mg/b) m
We integrate and solve for v :
(2-63)
v = - ?f (1 - e- bt/m ). (2-64)
36 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
We may obtain a formula useful for short times of fall by expanding the
exponential function in a power series :
v = -gt + \ % t 2 + ■ ■ ■ . (2-65)
Thus for a short time (t <£ m/b), v = —gt, approximately, and the effect
of air resistance can be neglected. After a long time, we see from Eq. (2-64)
that
v = ~ > if t » -r- •
b o
The velocity mg/b is called the terminal velocity of the falling body in ques-
tion. The body reaches within 1/e of its terminal velocity in a time
t = m/b. We could use the experimentally determined terminal velocity
to find the constant b. We now integrate Eq. (2-64), taking x — 0:
b z \ m /
By expanding the exponential function in a power series, we obtain
If t « m/b, x = — \qt 2 , as in Eq. (1-29). When t » m/b,
This result is easily interpreted in terms of terminal velocity. Why is
the positive constant present?
For small heavy bodies with large terminal velocities, a better approxi-
mation may be
F = bv 2 . (2-68)
The reader should be able to show that with the frictional force given by
Eq. (2-68), the result (taking x = v = at t = 0) is
(2-69)
x = -%gt 2 + i % t 3 + ■ ■ ■ . (2-67)
771/
v = -
-^«"(VS')
\7Yl
-gt, if *«^'
.-# i! «»>!■
2-6]
FALLING BODIES
r In cosh
b
m /
37
(2-70)
-&t 2 ,
if
t«
W
m
In 2 -
t,
if
f »
fm
w
Again there is a terminal velocity, given this time by (mg/b) 112 . The ter-
minal velocity can always be found as the velocity at which the frictional
force equals the gravitational force, and will exist whenever the frictional
force becomes sufficiently large at high velocities.
In the case of bodies falling from a great height, the variation of the
gravitational force with height should be taken into account. In this case,
we neglect air resistance, and measure x from the center of the earth.
Then if M is the mass of the earth and m the mass of the falling body,
F =
and
V(x) =
mMG
Fdx = -
mMG
(2-71)
(2-72)
where we have taken x s = oo in order to avoid a constant term in V(x).
Equation (2-4:5) becomes
12
(2-73)
-i^vio^r
The plus sign refers to ascending motion, the minus sign to descending
motion.
The function V(x) is plotted in v{x)
Fig. 2-2. We see that there are two
types of motion, depending on
whether E is positive or negative.
When E is positive, there is no turn-
ing point, and if the body is initially
moving upward, it will continue to
move upward forever, with decreas-
ing velocity, approaching the limit-
ing velocity
(OP
Vl = \ ~m ' ( 2_74) Fig. 2-2. Plot of V(x) = -(mMG/x).
38 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
When E is negative, there is a turning point at a height
mMG , n __.
x T = —^ ■ (2-75)
If the body is initially moving upward, it will come to a stop at xt, and
fall back to the earth. The dividing case between these two types of
motion occurs when the initial position and velocity are such that E = 0.
The turning point is then at infinity, and the body moves upward forever,
approaching the limiting velocity vi = 0. If E = 0, then at any height x,
the velocity will be
»• = >/—• (2_76)
This is. called the escape velocity for a body at distance x from the center
of the earth, because a body moving upward at height x with velocity v e
will just have sufficient energy to travel upward indefinitely (if there is
no air resistance).
To find x(t), we must evaluate the integral
dx /2 t, (2-77)
(* + =F)"
1/2 -\ m
where x is the height at t = 0. To solve for the case when E is negative,
we substitute
cos *=VH- < 2 - 78)
Equation (2-77) then becomes
mMG
(-E)-
\z / 2 cos 2 Odd = J-t. (2-79)
(We choose a positive sign for the integrand so that will increase when t
increases.) We can, without loss of generality, take x to be at the turning
point x T , since the body will at some time in its past or future career pass
through x T if no force except gravity acts upon it, provided E < 0. Then
d = 0, and
mMG , ,. a , [2 .
(0 + sin 6 cos 0) = ~ — t,
(—E) 3 I 2 \m
or
+ isin20= x F^t, (2-80)
V x T
2-7]
THE SIMPLE HARMONIC OSCILLATOR
39
and
Xt COS 0.
(2-81)
This pair of equations cannot be solved explicitly for x(t). A numerical
solution can be obtained by choosing a sequence of values of and finding
the corresponding values of x and t from Eqs. (2-80) and (2-81). That
part of the motion for which x is less than the radius of the earth will, of
course, not be correctly given, since Eq. (2-71) assumes all the mass of the
earth concentrated at x = (not to mention the fact that we have omitted
from our equation of motion the forces which would act on the body when
it collides with the earth).
The solution can be obtained in a similar way for the cases when E is
positive or zero.
2-7 The simple harmonic oscillator. The most important problem in
one-dimensional motion, and fortunately one of the easiest to solve, is the
harmonic or linear oscillator. The simplest example is that of a mass m
fastened to a spring whose constant is k. If we measure x from the re-
laxed position of the spring, then the spring exerts a restoring force
F = —kx.
(2-82)
The potential energy associated
with this force is
V(x) = ikx 2 . (2-83)
The equation of motion, if we
assume no other force acts, is
m^~ + kx = 0. (2-84)
-v.Q.QQCLCLP-
m
aa
Fig. 2-3. Model of a simple har-
monic oscillator.
Equation (2-84) describes the free harmonic oscillator. Its solution was
obtained in Section 2-5. The motion is a simple sinusoidal oscillation
about the point of equilibrium. In all physical cases there will be some
frictional force acting, though it may often be very small. As a good
approximation in most cases, particularly when the friction is small, we
can assume that the frictional force is proportional to the velocity. Since
this is the only kind of frictional force for which the problem can easily
be solved, we shall restrict our attention to this case. If we use Eq. (2-31)
for the frictional force with n = 1, the equation of motion then becomes
d 2 x , dx . , n
m dP +b dt +kx = -
(2-85)
This equation describes the damped harmonic oscillator. Its motion, at
least for small damping, consists of a sinusoidal oscillation of gradually
40 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
decreasing amplitude, as we shall show later. If the oscillator is subject
to an additional impressed force F(t), its motion will be given by
m W + b Tt + kx = F(tl (2_86)
If F(t) is a sinusoidally varying force, Eq. (2-86) leads to the phenomenon
of resonance, where the amplitude of oscillation becomes very large when
the frequency of the impressed force equals the natural frequency of the
free oscillator.
The importance of the harmonic oscillator problem lies in the fact that
equations of the same form as Eqs. (2-84)-(2-86) turn up in a wide variety
of physical problems. In almost every case of one-dimensional motion
where the potential energy function V(x) has one or more minima, the
motion of the particle for small oscillations about the minimum point fol-
lows Eq. (2-84). To show this, let V(x) have a minimum at x = x , and
expand the function V(x) in a Taylor series about this point:
V(x) = V(x ) + (^\ { x - x ) + i(g\ o (* - *o) 2
+ *(€?) (*-*o) 3 +---. (2-87)
'\dx S J:
*0
The constant V(x ) can be dropped without affecting the physical results.
Since x is a minimum point,
Making the abbreviations
* = (S). - <2 " 89)
x' = x — x 0) (2-90)
we can write the potential function in the form
V(x') = %kx' 2 H . (2-91)
For sufficiently small values of x', provided k ^ 0, we can neglect the terms
represented by dots, and Eq. (2-91) becomes identical with Eq. (2-83).
Hence, for small oscillations about any potential minimum, except in the
exceptional case k = 0, the motion is that of a harmonic oscillator.
When a solid is deformed, it resists the deformation with a force propor-
tional to the amount of deformation, provided the deformation is not too
2-8] LINEAR DIFFERENTIAL EQUATIONS 41
great. This statement is called Hooke's law. It follows from the fact that
the undeformed solid is at a potential-energy minimum and that the
potential energy may be expanded in a Taylor series in the coordinate
describing the deformation. If a solid is deformed beyond a certain point,
called its elastic limit, it will remain permanently deformed; that is, its
structure is altered so that its undeformed shape for minimum potential
energy is changed. It turns out in most cases that the higher-order terms
in the series (2-91) are negligible almost up to the elastic limit, so that
Hooke's law holds almost up to the elastic limit. When the elastic limit
is exceeded and plastic flow takes place, the forces depend in a complicated
way not only on the shape of the material, but also on the velocity of de-
formation and even on its previous history, so that the forces can no
longer be specified in terms of a potential-energy function.
Thus practically any problem involving mechanical vibrations reduces
to that of the harmonic oscillator at small amplitudes of vibration, that is,
so long as the elastic limits of the materials involved are not exceeded.
The motions of stretched strings and membranes, and of sound vibrations
in an enclosed gas or in a solid, result in a number of so-called normal
modes of vibration, each mode behaving in many ways like an independent
harmonic oscillator. An electric circuit containing inductance L, resist-
ance R, and capacitance C in series, and subject to an applied electro-
motive force E{t), satisfies the equation
where q is the charge on the condenser and dq/dt is the current. This
equation is identical in form with Eq. (2-86). Early work on electrical
circuits was often carried out by analogy with the corresponding mechani-
cal problem. Today the situation is often reversed, and the mechanical
and acoustical engineers are able to make use of the simple and effective
methods developed by electrical engineers for handling vibration prob-
lems. The theory of electrical oscillations in a transmission line or in a
cavity is similar mathematically to the problem of the vibrating string or
resonating air cavity. The quantum-mechanical theory of an atom can
be put in a form which is identical mathematically with the theory of a
system of harmonic oscillators.
2-8 Linear differential equations with constant coefficients. Equa-
tions (2-84)-(2-86) are examples of second-order linear differential equa-
tions. The order of a differential equation is the order of the highest
derivative that occurs in it. Most equations of mechanics are of second
order. (Why?) A linear differential equation is one in which there are
42 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
no terms of higher than first degree in the dependent variable (in this
case x) and its derivatives. Thus the most general type of linear differ-
ential equation of order n would be
a n (t) ^f + a„_i(«) gj^ + • • • + ai(t) g + a {t)x = b(t). (2-93)
If b(t) = 0, the equation is said to be homogeneous; otherwise it is inhomo-
geneous. Linear equations are important because there are simple general
methods for solving them, particularly when the coefficients a , a\, . . . , a n
are constants, as in Eqs. (2-84)-(2-86). In the present section, we shall
solve the problem of the free harmonic oscillator [Eq. (2-84)], and at the
same time develop a general method of solving any linear homogeneous
differential equation with constant coefficients. This method is applied
in Section 2-9 to the damped harmonic oscillator equation (2-85). In
Section 2-10 we shall study the behavior of a harmonic oscillator under a
sinusoidally oscillating impressed force. In Section 2-11 a theorem is
developed which forms the basis for attacking Eq. (2-86) with any im-
pressed force F(t), and the methods of attack are discussed briefly.
The solution of Eq. (2-84), which we obtained in Section 2-5, we now
write in the form
x = A sin (o) t + 6), w = Vk/m . (2-94)
This solution depends on two "arbitrary" constants A and 6. They are
called arbitrary because no matter what values are given to them, the
solution (2-94) will satisfy Eq. (2-84). They are not arbitrary in a phys-
ical problem, but depend on the initial conditions. It can be shown that
the general solution of any second-order differential equation depends on
two arbitrary constants. By this we mean that we can write the solution
in the form
x = x(t; C u C 2 ), (2-95)
such that for every value of C\ and C2, or every value within a certain
range, x(t; C\, C%) satisfies the equation and, furthermore, practically
every solution of the equation is included in the function x{t; C\, C2) for
some value of Ci and C 2 .* If we can find a solution containing two arbi-
trary constants which satisfies a second-order differential equation, then
we can be sure that practically every solution will be included in it. The
methods of solution of the differential equations studied in previous sec-
tions have all been such as to lead directly to a solution corresponding to
* The only exceptions are certain "singular" solutions which may occur in
regions where the mathematical conditions for a unique solution (Section 2-2)
are not satisfied.
2-8] LINEAR DIFFERENTIAL EQUATIONS 43
the initial conditions of the physical problem. In the present and subse-
quent sections of this chapter, we shall consider methods which lead to
a general solution containing two arbitrary constants. These constants
must then be given the proper values to fit the initial conditions of the
physical problem; the fact that a solution with two arbitrary constants
is the general solution guarantees that we can always satisfy the initial
conditions by proper choice of the constants.
We now state two theorems regarding linear homogeneous differential
equations:
Theorem I. If x = x x (t) is any solution of a linear homogeneous differ-
ential equation, and C is any constant, then x = Cx^t) is also a solution.
Theorem II. // x = Xi(t) and x = x 2 (t) are solutions of a linear homo-
geneous differential equation, then x = Xi(t) + x 2 (t) is also a solution.
We prove these theorems only for the case of a second-order equation,
since mechanical equations are generally of this type:
a 2 (t) § + a 1 (t) g + a (t)x = 0. (2-96)
Assume that x = x x (i) satisfies Eq. (2-96). Then
a 2 (t) ^fel + ai (t) ^ + aoWCxt) =
c [ a2(t) S 1 + fllW lit + a °® x i\ = °-
Hence x = Cx^t) also satisfies Eq. (2-96). If x x {t) and x 2 (t) both satisfy
Eq. (2-96), then
,.v d 2 (Xt + X 2 ) d(X! + x 2 )
zW <P r ai{t) - h a {t)(xi + x 2 )
+ [ aa(i) %- + fll(<) W + a °® x >] = °-
Hence x = x^t) + x 2 (t) also satisfies Eq. (2-96). The problem of finding
the general solution of Eq. (2-96) thus reduces to that of finding any
two independent "particular" solutions x^t) and x 2 {t), for then Theorems I
and II guarantee that
x = Ctx^t) + C 2 x 2 (t) (2-97)
44 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
is also a solution. Since this solution contains two arbitrary constants,
it must be the general solution. The requirement that xi(t) and x 2 (t)
be independent means in this case that one is not a multiple of the other.
If xi(<) were a constant multiple of x 2 (t), then Eq. (2-97) would really
contain only one arbitrary constant. The right member of Eq. (2-97)
is called a linear combination of xi and x 2 .
In the case of equations like (2-84) and (2-85), where the coefficients
are constant, a solution of the form x = e pt always exists. To show this,
assume that a , a lt and a 2 are all constant in Eq. (2-96) and substitute
x = e > H = pe ' W = v e ■ (2_98)
We then have
(a 2 p 2 + a lP + a )e pt = 0. (2-99)
Canceling out e pt , we have an algebraic equation of second degree in p.
Such an equation has, in general, two roots. If they are different, this
gives two independent functions e pt satisfying Eq. (2-96) and our prob-
lem is solved. If the two roots for p should be equal, we have found only
one solution, but then, as we shall show in the next section, the function
x = te pt (2-100)
also satisfies the differential equation. The linear homogeneous equation
of nth order with constant coefficients can also be solved by this method.
Let us apply the method to Eq. (2-84). Making the substitution (2-98),
we have
mp 2 + k = 0, (2-101)
whose solution is
p = ± yP± = ±ia, , « = ^ • (2-102)
This gives, as the general solution,
x = de^ot _|_ Cze-^ot. (2-103)
In order to interpret this result, we remember that
e*» = cos 6 + * sin 0. (2-104)
If we allow complex numbers x as solutions of the differential equation,
then the arbitrary constants Ci and C 2 must also be complex in order for
Eq. (2-103) to be the general solution. The solution of the physical
problem must be real, hence we must choose Ci and C 2 so that x turns out
2-8] LINEAR DIFFERENTIAL EQUATIONS 45
to be real. The sum of two complex numbers is real if one is the complex
conjugate of the other. If
C = a + ib, (2-105)
and
C* = a- ib, (2-106)
then
C + C* = 2a, C -C* = 2ib. (2-107)
Now e ia <>t i s the complex conjugate of e -iw °', so that if we set d = C,
C 2 = C*, then x will be real:
x = Ce iaot + C*e~ iu °K (2-108)
We could evaluate x by using Eqs. (2-104), (2-105), and (2-106), but
the algebra is simpler if we make use of the polar representation of a
complex number:
C = a + ib = re { °, (2-109)
C* = a — ib = re~ ie , (2-110)
where
r= (a 2 + b 2 ) m , tan0 = ^, (2-111)
a = r cos 0, b = r sin 8. (2-112)
The reader should verify that these equations follow algebraically from
Eq. (2-104). If we represent C as a point in the complex plane, then a
and b are its rectangular coordinates, and r and are its polar coordinates.
Using the polar representation of C, Eq. (2-108) becomes (we set r = %A)
= A cos (u t + 0). (2-113)
This is the general real solution of Eq. (2-84). It differs from the solu-
tion (2-94) only by a shift of ir/2 in the phase constant 6.
Setting B x = A cos 0, B 2 = —A sin 0, we can write our solution in
another form:
x = Bi cos u t + B 2 sin w <. (2-114)
The constants A, 0, or B u B 2 , are to be obtained in terms of the initial
values x Q , v at t = by setting
x = A cos = B u (2-115)
v = —o) A sin = u B 2 . (2-116)
A = U + &
r
\ <»l)
r
tan 0= "° ■ ,
a; wo
-Bi = x ,
B 2 = ^-
w
46 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
The solutions are easily obtained:
/ „.2\l/2
(2-117)
(2-118)
or
(2-119)
(2-120)
Another way of handling Eq. (2-103) would be to notice that, since
Eq. (2-84) contains only real coefficients, a complex function can satisfy
it only if both real and imaginary parts satisfy it separately. (The proof
of this statement is a matter of substituting x = u + iw and carrying
out a little algebra.) Hence if a solution is (we set r = A)
x = Ce iWot = Ae i{aat+6)
= A cos (u < + 0) + iA sin (u t + 9), (2-121)
then both the real and imaginary parts of this solution must separately be
solutions, and we have either solution (2-113) or (2-94). We can carry
through the solutions of linear equations like this, and perform any alge-
braic operations we please on them in their complex form (so long as we
do not multiply two complex numbers together), with the understanding
that at each step what we are really concerned with is only the real part
or only the imaginary part. This procedure is often useful in the treat-
ment of problems involving harmonic oscillations, and we shall use it in
Section 2-10.
It is often very convenient to represent a sinusoidal function as a com-
plex exponential :
cos = real part of e 10 = ~ ' (2-122)
it —it
sin 6 = imaginary part of e te = ^ (2-123)
Exponential functions are easier to handle algebraically than sines and
cosines. The reader will find the relations (2-122), (2-123), and (2-104)
useful in deriving trigonometric formulas. The power series for the sine
and cosine functions are readily obtained by expanding e l ° in a power series
and separating the real and imaginary parts. The trigonometric rule for
sin (A + B) and cos (A + B) can be easily obtained from the algebraic
rule for adding exponents. Many other examples could be cited.
2-9] THE DAMPED HARMONIC OSCILLATOR 47
2-9 The damped harmonic oscillator. The equation of motion for a
particle subject to a linear restoring force and a frictional force proportional
to its velocity is [Eq. (2-85)]
mx + bx + kx = 0, (2-124)
where the dots stand for time derivatives. Applying the method of Sec-
tion 2-8, we make the substitution (2-98) and obtain
mp 2 + bp + k = 0. (2-125)
The solution is
1/2
(2-126)
= _ _ + ±Y - a
P 1m l\2mj m.
[(,
We distinguish three cases: (a) k/m > (6/2m) 2 , (b) k/m < (b/2m) 2 , and
(c) k/m = (6/2m) 2 .
In case (a), we make the substitutions
"o ,= yj~ - (2-127)
T = 2^' (2- 12 8)
"i = («o - y 2 V 12 , (2-129)
where 7 is called the damping coefficient and (w /27r) is the natural fre-
quency of the undamped oscillator. There are now two solutions for p:
p = — J ± iui. (2-130)
The general solution of the differential equation is therefore
x = C 1 e- /t+iait + C 2 e- yt - iWlt . (2-131)
Setting
d = Ue ie , C 2 = Ue~ U , (2-132)
we have
x = Ae~ yt cos (»i« + 0). (2-133)
This corresponds to an oscillation of frequency (w!/27r) with an amplitude
Ae- yt which decreases exponentially with time (Fig. 2-4). The constants
A and depend upon the initial conditions. The frequency of oscillation
is less than without damping. The solution (2-133) can also be written
x = e~ yt (Bi cosgM + B 2 sinwiO- (2-134)
48
MOTION OF A PARTICLE IN ONE DIMENSION
[CHAP. 2
Fig. 2-4. Motion of damped harmonic oscillator. Heavy curve: x =
Ae~ yt cos ut, 7 = w/8. Light curve: a; = ±Ae~ yt .
In terms of the constants co and 7, Eq. (2-124) can be written
x + 2lx + v>%x = 0. (2-135)
This form of the equation is often used in discussing mechanical oscilla-
tions.
The total energy of the oscillator is
E = hnx 2 + Jfcc 2 .
(2-136)
In the important case of small damping, 7 <C «o, we can set o>i = o>o and
neglect 7 compared with w , and we have for the energy corresponding to
the solution (2-133), approximately,
E = ikA 2 e~ 2yt = E e- 2yt .
(2-137)
Thus the energy falls off exponentially at twice the rate at which the ampli-
tude decays. The fractional rate of decline or logarithmic derivative of E is
I §E
E dt
dlnE
dt
= —27.
(2-138)
We now consider case (b), (co < 7). In this case, the two solutions
for p are
2-9] THE DAMPED HARMONIC OSCILLATOR 49
p=-y 1 = -y- (7 2 - cog) 1 ' 2 ,
v = -i% = -v + (t 2 - «S) 1/a .
The general solution is
(2-139)
x = C ie ~ yit + C 2 e- y2t . (2-140)
These two terms both decline exponentially with time, one at a faster rate
than the other. The constants C\ and C 2 may be chosen to fit the initial
conditions. The reader should determine them for two important cases:
x ^ 0, v = and x = 0, v ?± 0, and draw curves x{f) for the two
cases.
In case (c), (co = T), we have only one solution for p:
P = —y. (2-141)
The corresponding solution for x is
* = e~ yt . (2-142)
We now show that, in this case, another solution is
x = te~ yt . (2-143)
To prove this, we compute
x = e~ yt - yte~ y \
(2-144)
* = -2Te- 7J + y 2 te~ yt .
The left side of Eq. (2-135) is, for this x,
x + 27^ + o%x = (cwg — y 2 )te~ yt . (2-145)
This is zero if w = 7. Hence the general solution in case co = 7 is
x = (C x + C 2 t)e~ n . (2-146)
This function declines exponentially with time at a rate intermediate be-
tween that of the two exponential terms in Eq. (2-140):
?i > y > 7 2 . (2-147)
Hence the solution (2-146) falls to zero faster after a sufficiently long time
than the solution (2-140), except in the case C 2 = in Eq. (2-140). Cases
(a), (b), and (c) are important in problems involving mechanisms which
approach an equilibrium position under the action of a frictional damping
50
MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
Fig. 2-5. Return of harmonic oscillator to equilibrium, (a) Underdamped.
(b) Overdamped. (o) Critically damped.
force, e.g., pointer reading meters, hydraulic and pneumatic spring returns
for doors, etc. In most cases, it is desired that the mechanism move
quickly and smoothly to its equilibrium position. For a given damping
coefficient 7, or for a given u , this is accomplished in the shortest time
without overshoot if co = 7 [case (c)]. This case is called critical damping.
If «o < 7, the system is said to be overdamped; it behaves sluggishly and
does not return as quickly to x = as for critical damping. If w > 7,
the system is said to be underdamped; the coordinate x then overshoots the
value x = and oscillates. Note that at critical damping, ui = 0, so
that the period of oscillation becomes infinite. The behavior is shown in
Fig. 2-5 for the case of a system displaced from equilibrium and released
(#0 n* 0, v = 0). The reader should draw similar curves for the case
where the system is given a sharp blow at t = (i.e., x = 0, v j£ 0).
2-10 The forced harmonic oscillator. The harmonic oscillator subject
to an external applied force is governed by Eq. (2-86). In order to sim-
plify the problem of solving this equation, we state the following theorem :
Theorem III. // x»(<) is a solution of an inhomogeneous linear equation
[e.g., Eq. (2-86)], and Xh(t) is a solution of the corresponding homogeneous
equation [e.g., Eq. (2-85)], then x(t) = Xi(t) + Xh{t) is also a solution of
the inhomogeneous equation.
This theorem applies whether the coefficients in the equation are constants
or functions of t. The proof is a matter of straightforward substitution,
and is left to the reader. In consequence of Theorem III, if we know the
general solution Xh of the homogeneous equation (2-85) (we found this in
Section 2-9), then we need find only one particular solution Xi of the in-
homogeneous equation (2-86). For we can add Xi to Xh and obtain a solu-
tion of Eq. (2-86) which contains two arbitrary constants and is therefore
the general solution.
2-10] THE FORCED HARMONIC OSCILLATOR 51
The most important case is that of a sinusoidally oscillating applied
force. If the applied force oscillates with angular frequency w and ampli-
tude F , the equation of motion is
d x dx
m -£p + b -£ + kx = F cos (at + O ), (2-148)
where 6 is a constant specifying the phase of the applied force. There are,
of course, many solutions of Eq. (2-148), of which we need find only one.
From physical considerations, we expect that one solution will be a steady
oscillation of the coordinate x at the same frequency as the applied force:
x = A a cos (at + 0.). (2-149)
The amplitude A, and phase S of the oscillations in x will have to be de-
termined by substituting Eq. (2-149) in Eq. (2-148). This procedure is
straightforward and leads to the correct answer. The algebra is simpler,
however, if we write the force as the real part of a complex function:*
F(t) = Re(F e iat ), (2-150)
F = Foe"". (2-151)
Thus if we can find a solution x(t) of
m ll& + b 7fi + kx= F ° e """' (2-152)
then, by splitting the equation into real and imaginary parts, we can show
that the real part of x(t) will satisfy Eq. (2-148). We assume a solution of
the form
ico t
x = x e ,
so that
x = twxoe*", x = -w 2 x e iw '. (2-153)
Substituting in Eq. (2-152), we solve for x :
Xo = _^ (2 _ 154)
wo — w + 2iTco
The solution of Eq. (2-152) is therefore
x = x e-= 2 ^f^ • (2-155)
Wo — w 2 + 2tTw
* Note the use of roman type (F, x) to distinguish complex quantities from
the corresponding real quantities {F, x).
52 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
We are often more interested in the velocity
t'coFo
iut
±=^ 1 1— T— • (2-156)
TO w — w + 2i7u
The simplest way to write Eq. (2-156) is to express all complex factors in
polar form [Eq. (2-109)] :
i = e" /2 , (2-157)
co 2 , - co 2 + 2i7co = [(col - co 2 ) 2 + 4Y 2 co 2 ] 1/2 exp ' ' - 1 27w
I i tan x
2 2
COq — w
If we use these expressions, Eq. (2-156) becomes
(2-158)
where
" m[(co 2 - co 2 ) 2 + 47V] 1 ' 2 ' ( 59)
= I - tan" 1 -^- = tan" 1 »L=L«? , (2 -l 6 0)
2 «o — w 2 27co
^=[^yfky^ (2 " 161)
By Eq. (2-159),
x = Re(x)
F co
m [(cog - co 2 ) 2 + 47V]
and
a; = Re(x) = Re (k/ico)
F 1
— cos (cot + <?o + (3), (2-163)
m [(a>i — a*)' + 47 2 co 2 ] 1/2
This is a particular solution of Eq. (2-148) containing no arbitrary con-
stants. By Theorem III and Eq. (2-133), the general solution (for the
underdamped oscillator) is
' = Ae ~" oos <-' + «> + m _ Jf + 47vi-» *" *-+*>+»•
(2-165)
2-10] THE FORCED HARMONIC OSCILLATOR 53
This solution contains two arbitrary constants A, 6, whose values are de-
termined by the initial values x , v at t — 0. The first term dies out ex-
ponentially in time and is called the transient. The second term is called
the steady state, and oscillates with constant amplitude. The transient de-
pends on the initial conditions. The steady state which remains after the
transient dies away is independent of the initial conditions.
In the steady state, the rate at which work is done on the oscillator by
the applied force is
F 2
xF(t) = -2 — " cos (cot + e ) cos (cot + B + 0)
m [(co 2 — a>o) + 4tV] 1/2
_ F% co cos /3 cos 2 (cot + fl ) F% w sin sin 2(a>t + O )
m [(a, 2 - w 2 ) 2 + 4Y 2 w 2 ] 1/2 2m [(co 2 - co 2 ) 2 + 4Y 2 co 2 ] 1/2
(2-166)
The last term on the right is zero on the average, while the average value
of cos 2 (tot + O ) over a complete cycle is \. Hence the average power de-
livered by the applied force is
Pav = <**«»„ = ^^ " — - , (2-167)
2m [(to 2 - col) 2 + 47V] 1/2
or
-Pav = %FoX m cos /3, (2-168)
where x m is the maximum value of x. A similar relation holds for power
delivered to an electrical circuit. The factor cos ft is called the power
factor. In the electrical case, is the phase angle between the current and
the applied emf. Using formula (2-162) for cos/3, we can rewrite Eq.
(2-167) :
F 2 yco 2
m (co 2 — cog) 2 + 4T 2 co 2
It is easy to show that in the steady state power is supplied to the oscillator
at the same average rate that power is being dissipated by friction, as of
course it must be. The power P av has a maximum for co ±= co . In Fig.
2-6, the power P av (in arbitrary units) and the phase of ff of steady-state
forced oscillations are plotted against co for two values of 7. The heavy
curves are for small damping; the light curves are for greater damping.
Formula (2-169) can be simplified somewhat in case y <JC co . In this case,
P av is large only near the resonant frequency co , and we shall deduce a
formula valid near co = co . Defining
Aw = co — w , (2-170)
54
MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
■P»v, (7 = J«o)
-x/2
Fig. 2-6. Power and phase of forced harmonic oscillations.
and assuming Aw « w , we have
(w 2 — wo) = (w + w ) Aw = 2w Aw,
Hence
(2-171)
(2-172)
(2-173)
This simple formula gives a good approximation to P av near resonance.
The corresponding formula for /3 is
—Aw
w 2
=
2
w .
->
=
Fl
4m
7
av
(Aw)
2 _j_ 7 2
cos |3 =
sin j3 =
[(Aw) 2 + T2]i/2 ' "*" >- " [(Aw) 2 + Y 2 ] 1 / 2
When w « w , = ir/2, and Eq. (2-164) becomes
x +■■%>- cob (at +e ) = nr"
worn K
(2-174)
(2-175)
This result is easily interpreted physically; when the force varies slowly,
the particle moves in such a way that the applied force is just balanced by
the restoring force. When w » w , /3 = —tt/2, and Eq. (2-164) becomes
x= - -4 s - cos (jut + 6 ) = ~^--
w z m w 2 m
(2-176)
2-10] THE FORCED HARMONIC OSCILLATOR 55
The motion now depends only on the mass of the particle and on the fre-
quency of the applied force, and is independent of the friction and the
restoring force. This result is, in fact, identical with that obtained in Sec-
tion 2-3 [see Eqs. (2-15) and (2-19)] for a free particle subject to an
oscillating force.
We can apply the result (2-165) to the case of an electron bound to an
equilibrium position x = by an elastic restoring force, and subject to an
oscillating electric field :
E x = E cos tot, (2-177)
F = — eE cos ut. (2-178)
The motion will be given by
x = Ac'" cos ( Wl < + 6) - <%> sin <■<* + A
m [(co 2 - cog) 2 + 47 2 co 2 ] 1/2
(2-179)
The term of interest here is the second one, which is independent of the
initial conditions and oscillates with the frequency of the electric field.
Expanding the second term, we get
„ _ eE sin ff cos wt eE cos /3 sin cot
;£ — — - - - - —
tn [(co 2 - cog) 2 + 47 2 co 2 ] 1/2 m [(co 2 - co 2 ) 2 + 4T 2 co 2 ] 1/2
2
-eE cos ut coq
co
tn [(co 2 - co 2 ) 2 + 47 V]
eE sin ut 2Tco
tn [(co 2 - co 2 ) 2 + 4T 2 co 2 ]
(2-180)
The first term represents an oscillation of x in phase with the applied force
at low frequencies, 180° out of phase at high frequencies. The second term
represents an oscillation of x that is 90° out of phase with the applied force,
the velocity x for this term being in phase with the applied force. Hence
the second term corresponds to an absorption of energy from the applied
force. The second term contains a factor 7 and is therefore small, if
7 <C co , except near resonance. If we imagine a dielectric medium con-
sisting of electrons bound by elastic forces to positions of equilibrium, then
the first term in Eq. (2-180) will represent an electric polarization propor-
tional to the applied oscillating electric field, while the second term will
represent an absorption of energy from the electric field. Near the resonant
frequency, the dielectric medium will absorb energy, and will be opaque
to electromagnetic radiation. Above the resonant frequency, the dis-
placement of the electrons is out of phase with the applied force, and the
56 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
resulting electric polarization will be out of phase with the applied electric
field. The dielectric constant and index of refraction will be less than
one. For very high frequencies, the first term of Eq. (2-180) approaches
the last term of Eq. (2-18), and the electrons behave as if they were free.
Below the resonant frequency, the electric polarization will be in phase
with the applied electric field, and the dielectric constant and index of re-
fraction will be greater than one.
Computing the dielectric constant from the first term in Eq. (2-180), in the
same manner as for a free electron [see Eqs. (2-20)-(2-26)], we find, for N elec-
trons per unit volume:
A-wNe cop — to
€=1 + 71 2,2 ,^22 " < 2 - 181 >
m (o>o — «■ ) + 47 to
The index of refraction for electromagnetic waves (n = 1) is
c , ,1/2 1/2 ,~ 100 ,
n = - = (tie) = e . (2-182)
v
For very high or very low frequencies, Eq. (2-181) becomes
6=1 + ^f- , co « oo, (2-183)
rruoQ
2
TOO) 2
4irNe
e = 1 - ^-|- , fa) » coo. (2-184)
The mean rate of energy absorption per unit volume is given by Eq. (2-169):
- = ^ 2 l\ 2 2 - (S-185)
dt m (a) — coo) + 47 fa>
The resulting dielectric constant and energy absorption versus frequency
are plotted in Fig. 2-7. Thus the dielectric constant is constant and
greater than one at low frequencies, increases as we approach the resonant
frequency, falls to less than one in the region of "anomalous dispersion"
where there is strong absorption of electromagnetic radiation, and then
rises, approaching one at high frequencies. The index of refraction will
follow a similar curve. This is precisely the sort of behavior which is
exhibited by matter in all forms. Glass, for example, has a constant dielec-
tric constant at low frequencies; in the region of visible light its index of
refraction increases with frequency; and it becomes opaque in a certain
2-10] THE FORCED HARMONIC OSCILLATOR 57
Fig. 2-7. Dielectric constant and energy absorption for medium containing
harmonic oscillators.
band in the ultraviolet. X-rays are transmitted with an index of refrac-
tion very slightly less than one. A more realistic model of a transmitting
medium would result from assuming several different resonant frequencies
corresponding to electrons bound with various values of the spring con-
stant k. This picture is then capable of explaining most of the features in
the experimental curves for e or n vs. frequency. Not only is there qualita-
tive agreement, but the formulas (2-181 )-(2-185) agree quantitatively
with experimental results, provided the constants JV, to , and 7 are properly
chosen for each material. The success of this theory was one of the reasons
for the adoption, until the year 1913, of the "jelly model" of the atom, in
which electrons were imagined embedded in a positively charged jelly in
which they oscillated as harmonic oscillators. The experiments of Ruther-
ford in 1913 forced physicists to adopt the "planetary" model of the atom,
but this model was unable to explain even qualitatively the optical and
electromagnetic properties of matter until the advent of quantum mechan-
ics. The result of the quantum-mechanical treatment is that, for the inter-
action of matter and radiation, the simple oscillator picture gives essentially
correct results when the constants are properly chosen.*
We now consider an applied force F(t) which is large only during a
short time interval U and is zero or negligible at all other times. Such a
force is called an impulse, and corresponds to a sudden blow. We assume
the oscillator initially at rest at x = 0, and we assume the time St so short
that the mass moves only a negligibly small distance while the force is
acting. According to Eq. (2-4), the momentum just after the force is
applied will equal the impulse delivered by the force :
mvo
= Po = JF dt, (2-186)
where v is the velocity just after the impulse, and the integral is taken
over the time interval U during which the force acts. After the impulse,
* See John C. Slater, Quantum Theory of Matter. New York: McGraw-Hill
Book Co., 1951. (Page 378.)
58 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
the applied force is zero, and the oscillator must move according to Eq.
(2-133) if the damping is less than critical. We are assuming St so small
that the oscillator does not move appreciably during this time, hence we
choose = — (x/2) — ait , in order that x = at t = t , where t is
the instant at which the impulse occurs :
x = Ae~ yt sin [ai(t - t )]. (2-187)
The velocity at t = t is
v = aiAe"" . (2-188)
Thus
A = ^ e yt °. (2-189)
The solution when an impulse po is delivered at t ?= t to an oscillator at
rest is therefore
(0, * < t 0)
* = { _BL. e - T «-<o> sin [wi(< _ <o)] t > f (2-190)
Here we have neglected the short time 8t during which the force acts.
We see that the result of an impulse-type force depends only on the total
impulse po delivered, and is independent of the particular form of the func-
tion F(t), provided only that F(t) is negligible except during a very short
time interval St. Several possible forms of F(t) which have this property
are listed below:
(0, t < t ,
Po/St, t < t < t + St, (2-191)
0, t> t + St,
m = sW^ exp i~ (1 J^\' ~™ <t<cc (2 " 193)
The reader may verify that each of these functions is negligible except
within an interval of the order of St around t , and that the total impulse
delivered by each is p . The exact solution of Eq. (2-86) with F(f) given
by any of the above expressions must reduce to Eq. (2-190) when St — *
(see Problem 23).
2-11] THE PRINCIPLE OF SUPERPOSITION 59
2-11 The principle of superposition. Harmonic oscillator with arbi-
trary applied force. An important property of the harmonic oscillator is
that its motion x(t), when subject to an applied force F(t) which can be
regarded as the sum of two or more other forces Fi(t), F 2 {t), . . . , is the
sum of the motions x^t), x 2 (t), . . . , which it would have if each of the
forces F n {t) were acting separately. This principle applies to small mechan-
ical vibrations, electrical vibrations, sound waves, electromagnetic waves,
and all physical phenomena governed by linear differential equations. The
principle is expressed in the following theorem:
Theorem IV. Let the {finite or infinite*) set of functions x n (t), n = 1,2, 3,
. . . , be solutions of the equations
mx n + bx n + kx n = F n (t), (2-194)
and let
F(t) = X) *"»(*)• (2-195)
n
*(0 = Z) x «® (2-196)
n
mx + bx + kx = F(t). (2-197)
Then the function
satisfies the equation
To prove this theorem, we substitute Eq. (2-196) in the left side of Eq.
(2-197):
mx + bx + kx = m ^ x n + b ^ x n + k ^ x n
n n n
= ^ (mx n + bx n + kx n )
n
= !>»(*>
= F(t).
This theorem enables us to find a solution of Eq. (2-197) whenever the
force F(t) can be expressed as a sum of forces F n {t) for which the solutions
of the corresponding equations (2-194) can be found. In particular,
whenever F(t) can be written as a sum of sinusoidally oscillating terms :
^(0 = £ C " cos ("»* + "»)> (2-198)
* When the set of functions is infinite, there are certain mathematical restric-
tions which need not concern us here.
60 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
a particular solution of Eq. (2-197) will be, by Theorem IV and Eq. (2-164),
* = D — 771 * , , a 2ll/2 sin {uj + 9 n + ft,), (2-199)
V m [(u 2 - <4) + 47 2 co 2 ] 1/2
2 2
^ = tan ^^7"-
The general solution is then
a —~tt i . i a \ i v~* C« sin (co„£ + n + ft„) . .
x = Ae cos (wi< + 6) + 2*,— TTl 2x2 , , 2 2,1/2 ' C 2-200 )
» m [(co 2 - co 2 ,) 2 + 47 2 co 2 ] 1/2
where A and are, as usual, to be chosen to make the solution (2-200) fit
the initial conditions.
We can write Eqs. (2-198) and (2-199) in a different form by setting
A n = C n cos 0„, B n = — C n sin 6 n . (2-201)
Then
F(t) = J2 (A n cos u n t + B n sin wj), (2-202)
n
and
_ y-> A n sin (u n t + ff B ) — B n cos (co»< + ft>) f2-20*ri
V m[(co 2 - cog) 2 + 47 2 co 2 ] 1 ' 2 ' ^ ^
An important case of this kind is that of a periodic force F(t), that is, a
force such that
F(t + T) = F(t), (2-204)
where T is the period of the force. For any continuous function F(t) satis-
fying Eq. (2-204) (and, in fact, even for only piecewise continuous func-
tions), it can be shown that F(t) can always be written as a sum of sinus-
oidal functions:
where
F(t) = ±A +J1 (i B cos ^ + B n sin ^) , (2-205)
n = l
rT
An =Th
rT
F(t) cos ?^ dt, n = 0, 1, 2,
F(t) sin ^ <ft, n = 1, 2, 3,
(2-206)
This result enables us, at least in principle, to solve the problem of the
forced oscillator for any periodically varying force. The sum in Eq. (2-205)
2-11] THE PRINCIPLE OF SUPERPOSITION 61
is called a Fourier series* The actual computation of the solution by this
method is in most cases rather laborious, particularly the fitting of the
constants A, 6 in Eq. (2-200) to the initial conditions. However, the knowl-
edge that such a solution exists is often useful in itself. If any of the fre-
quencies 2irn/T coincides with the natural frequency w of the oscillator,
then the corresponding terms in the series in Eqs. (2-199) or (2-203) will
be relatively much larger than the rest. Thus a force which oscillates
nonsinusoidally at half the frequency w may cause the oscillator to per-
form a nearly sinusoidal oscillation at its natural frequency co .
A generalization of the Fourier series theorem [Eqs. (2-205) and (2-206)]
applicable to nonperiodic forces is the Fourier integral theorem, which
allows us to represent any continuous (or piecewise continuous) function
F(t), subject to certain limitations, as a superposition of harmonically
oscillating forces. By means of Fourier series and integrals, we may solve
Eq. (2-197) for almost any physically reasonable force F(t). We shall not
pursue the subject further here. Suffice it to say that while the methods
of Fourier series and Fourier integrals are of considerable practical value
in solving vibration problems, their greatest importance in physics probably
lies in the fact that in principle such a solution exists. Many important
results can be deduced without ever actually evaluating the series or
integrals at all.
A method of solution known as Green's method is based on the solution
(2-190) for an impulse-type force. We can think of any force F(t) as the
sum of a series of impulses, each acting during a short time St and deliver-
ing an impulse F(t) St:
F(t) = £ F n (t), (2-207)
n— — oo
0, if t < t n , where t n = n St,
Fn(t) = F(t n ), if t n < t < tn+u (2-208)
0, if t > tn +1 .
As St —> 0, the sum of all the impulse forces F n (t) will approach F(t). (See
Fig. 2-8.) According to Theorem IV and Eq. (2-190), a solution of
Eq. (2-197) for a force given by Eq. (2-207) is
* For a proof of the above statements and a more complete discussion of
Fourier series, see Dunham Jackson, Fourier Series and Orthogonal, Polynomials.
Menasha, Wisconsin: George Banta Pub. Co., 1941. (Chapter 1.)
62
MOTION OF A PARTICLE IN ONE DIMENSION
[CHAP. 2
F(t)
Fig. 2-8. Representation of a force as a sum of impulses. Heavy curve:
F(f). Light curve: £„F»(fl.
where t no < t < t no+1 . If we let St -» and write t„ = t', Eq. (2-209)
becomes
J — I
x(ty
ma)i
e- 7( '- (,) sin [ Wl (« - t')] dt'.
(2-210)
The function
G(t, t>) = \
0, if t' > t,
(2-211)
mu>\
sin[w!(< — t')], if t' < t,
is called the Green's function for Eq. (2-197). In terms of Green's function,
x (t) = f G(t, t')F(t') dt'. (2-212)
J 00
If the force F(t) is zero for t < t , then the solution (2-210) will give
x(t) = f or t < t - This solution is therefore already adjusted to fit the
initial condition that the oscillator be at rest before the application of the
force. For any other initial condition, a transient given by Eq. (2-133),
with appropriate values of A and 6, will have to be added. The solution
(2-210) is useful in studying the transient behavior of a mechanical sys-
tem or electrical circuit when subject to forces of various kinds.
Problems
1. A tug of war is held between two teams of five men each. Each man weighs
160 lb and can initially pull on the rope with a force of 200 lb-wt. At first the
teams are evenly matched, but as the men tire, the force with which each man
pulls decreases according to the formula
(200 lb-wt) e
-tlT
where the mean tiring time r is 10 sec for one team and 20 sec for the other.
Find the motion, (g — 32 ft-sec -2 .) What is the final velocity of the two
teams? Which of our assumptions is responsible for this unreasonable result?
PROBLEMS 63
2. A high-speed proton of electric charge e moves with constant speed v
in a straight line past an electron of mass m, charge — e, initially at rest. The
electron is at a distance a from the path of the proton, (a) Assume that the
proton passes so quickly that the electron does not have time to move appre-
ciably from its initial position until the proton is far away. Show that the
component of force in a direction perpendicular to the line along which the
proton moves is
2
F = 7~i~~i 2 2x3/2 ' (electrostatic or gaussian units)
(o + vot )
where a is the distance of the electron from the path of the proton and t =
when the proton passes closest to the electron, (b) Assume that the electron
moves only along a line perpendicular to the path of the proton. Find the final
kinetic energy of the electron, (c) Write the component of the force in a direc-
tion parallel to the proton velocity, and calculate the net impulse in that direc-
tion delivered to the electron. Does this justify the assumption in part (b)?
3. A particle which had originally a velocity vo is subject to a force given
by Eq. (2-192). (a) Find v(t) and x(t). (b) Show that as St -► 0, the motion
approaches motion at constant velocity with an abrupt change in velocity at
t = to of amount po/m.
4. A particle initially at rest is subject, beginning at < = 0, to a force
F = Foe~ 7( cos(cot+0).
(a) Find its motion, (b) How does the final velocity depend on 6, and on w?
[Hint: The algebra is simplified by writing cos (co«+ 0) in terms of complex expo-
nential functions.]
5. A boat with initial velocity vo is slowed by a frictional force
F = -be a \
(a) Find its motion, (b) Find the time and the distance required to stop.
6. A jet engine which develops a constant maximum thrust F is used to power
a plane with a frictional drag proportional to the square of the velocity. If the
plane starts at t = with a negligible velocity and accelerates with maximum
thrust, find its velocity v(t).
7. Find v(t) and x(t) for a particle of mass m which starts at xo = with
velocity v , subject to a force given by Eq. (2-31) with n ^ 1. Find the time
to stop, and the distance required to stop, and verify the remarks in the last
paragraph of Section 2-4.
8. (a) A body of mass m slides on a rough horizontal surface. The coefficient
of static friction is y. a , and the coefficient of sliding friction is y.. Devise an
analytic function F(v) to represent the frictional force which has the proper
constant value at appreciable velocities and reduces to the static value at very
low velocities, (b) Find the motion under the force you have devised if the
body starts with an initial velocity vo.
9. A particle of mass m is repelled from the origin by a force inversely pro-
portional to the cube of its distance from the origin. Set up and solve the equa-
tion of motion if the particle is initially at rest at a distance xq from the origin.
64
MOTION OF A PARTICLE IN ONE DIMENSION
[CHAP. 2
10. (a) A mass m is connected to the origin with a spring of constant k,
whose length when relaxed is I. The restoring force is very nearly proportional
to the amount the spring has been stretched or compressed so long as it is not
stretched or compressed very far. However, when the spring is compressed too
far, the force increases very rapidly, so that it is impossible to compress the
spring to less than half its relaxed length. When the spring is stretched more
than about twice its relaxed length, it begins to weaken, and the restoring
force becomes zero when it is stretched to very great lengths, (a) Devise a force
function F(x) which represents this behavior. (Of course a real spring is de-
formed if stretched too far, so that F becomes a function of its previous history,
but you are to assume here that F depends only on x.) (b) Find V(x) and
describe the types of motion which may occur.
11. A particle is subject to a force
x 3
(a) Find the potential V(x), describe the nature of the solutions, and find the
solution x(t). (b) Can you give a simple interpretation of the motion when
E 2 » hal
v
x)
' \
i
-Xi \
Xl
-V<\
Figure 2-9
12. An alpha particle in a nucleus is held by a potential having the shape
shown in Fig. 2-9. (a) Describe the kinds of motion that are possible, (b) De-
vise a function V(x) having this general form and having the values — Vo
and Vi at x = and x = ±x\, and find the corresponding force.
13. Derive the solutions (2-69) and (2-70) for a falling body subject to a
frictional force proportional to the square of the velocity.
14. A body of mass m falls from rest through a medium which exerts a fric-
tional drag fee" 1 »'. (a) Find its velocity v(t). (b) What is the terminal velocity?
(c) Expand your solution in a power series in t, keeping terms up to t 2 . (d) Why
does the solution fail to agree with Eq. (1-28) even for short times <?
15. A projectile is fired vertically upward with an initial velocity vo. Find its
motion, assuming a frictional drag proportional to the square of the velocity.
(Constant g.)
16. Derive equations analogous to Eqs. (2-80) and (2-81) for the mo-
tion of a body whose velocity is greater than the escape velocity. [Hint: Set
sinh/3 = (Ex/mMG) 1 ' 2 .]
17. Find the motion of a body projected upward from the earth with a velocity
equal to the escape velocity. Neglect air resistance.
PBOBLEMS 65
18. Find the general solution for the motion of a body subject to a linear re-
pelling force F = kx. Show that this is the type of motion to be expected in
the neighborhood of a point of unstable equilibrium.
19. The potential energy for the force between two atoms in a diatomic mole-
cule has the approximate form:
t// \ a . b
where x is the distance between the atoms and a, b are positive constants, (a)
Find the force, (b) Assuming one of the atoms is very heavy and remains at
rest while the other moves along a straight line, describe the possible motions,
(c) Find the equilibrium distance and the period of small oscillations about the
equilibrium position if the mass of the lighter atom is m.
20. A particle of mass m is subject to a force given by
F = B (£. - ^ 4- 21a% \
\x 2 is 1 " iSf
(a) Find and sketch the potential energy. (B and a are positive.) (b) Describe
the types of motion which may occur. Locate all equilibrium points and de-
termine the frequency of small oscillations about any which are stable, (c) A
particle starts at x = 3a/2 with a velocity v = —vo, where Do is positive. What is
the smallest value of wo for which the particle may eventually escape to a very
large distance? Describe the motion in that case. What is the maximum velocity
the particle will have? What velocity will it have when it is very far from its
starting point?
21. A particle of mass m moves in a potential well given by
. , _ -V a 2 (a 2 + x 2 )
VW 8a* + x* '
(a) Sketch V(x) and F(x). (b) Discuss the motions which may occur. Locate
all equilibrium points and determine the frequency of small oscillations about
any that are stable, (c) A particle starts at a great distance from the potential
well with velocity vo toward the well. As it passes the point x = a, it suffers a
collision with another particle, during which it loses a fraction a of its kinetic
energy. How large must a be in order that the particle thereafter remain trapped
in the well? How large must a be in order that the particle be trapped in one
side of the well? Find the turning points of the new motion if a = 1.
22. Starting with e 2i> = (e ie ) 2 , obtain formulas for sin 20, cos 20 in terms of
sin 0, cos 6.
23. Find the general solutions of the equations:
(a) mx + bx — kx = 0,
(b) mx — bx + kx = 0.
Discuss the physical interpretation of these equations and their solutions, assum-
ing that they are the equations of motion of a particle.
66 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2
24. Show that when co 2 , — T 2 is very small, the underdamped solution (2-133)
is approximately equal to the critically damped solution (2-146), for a short
time interval. What is the relation between the constants C\, C2 and A, 61 This
result suggests how one might discover the additional solution (2-143) in the
critical case.
25. A mass m subject to a linear restoring force —kx and damping — bx is dis-
placed a distance xo from equilibrium and released with zero initial velocity.
Find the motion in the underdamped, critically damped, and overdamped cases.
26. Solve Problem 25 for the case when the mass starts from its equilibrium
position with an initial velocity vo. Sketch the motion for the three cases.
27. A mass of 1000 kgm drops from a height of 10 m on a platform of negligible
mass. It is desired to design a spring and dashpot on which to mount the plat-
form so that the platform will settle to a new equilibrium position 0.2 m below
its original position as quickly as possible after the impact without overshooting.
(a) Find the spring constant k and the damping constant 6 of the dashpot.
Why does the result seem to contradict the remarks at the end of Section 2-9?
(b) Find, to two significant figures, the time required for the platform to settle
within 1 mm of its final position.
28. A force Fo(l — e~ at ) acts on a harmonic oscillator which is at rest at
t = 0. The mass is m, the spring constant k = ±ma 2 , and b = ma. Find the
motion. Sketch x(t).
*29. Solve Problem 28 for the case k = ma 2 , b = 2ma.
30. A force Fo cos (cot + #o) acts on a damped harmonic oscillator beginning
at t = 0. (a) What must be the initial values of x and v in order that there be
no transient? (b) If xo = vq = 0, find the amplitude A and phase 9 of the
transient in terms of Fo, 80.
31. An undamped harmonic oscillator of mass m, natural frequency coo, is
initially at rest and is subject at t = to a blow so that it starts from xo =
with initial velocity vo and oscillates freely until t = 3ir/2coo. From this time
on, a force F = B cos (cot + 0) is applied. Find the motion.
32. An underdamped harmonic oscillator is subject to an applied force
F = F e~ at cos (cot + 0).
Find a particular solution by expressing F as the real part of a complex exponen-
tial function and looking for a solution for x having the same exponential time
dependence.
33. (a) Find the motion of a damped harmonic oscillator subject to a constant
applied force Fo, by guessing a "steady-state" solution of the inhomogeneous
equation (2-86) and adding a solution of the homogeneous equation, (b) Solve
the same problem by making the substitution x' = x — a, and choosing the
constant a so as to reduce the equation in x' to the homogeneous equation (2-85).
Hence show that the effect of the application of a constant force is merely to shift
the equilibrium position without affecting the nature of the oscillations.
* An asterisk is used, as explained in the Preface, to indicate problems which
may be particularly difficult.
PROBLEMS 67
34. Find the motion of a mass m subject to a restoring force — kx, and to a
damping force {±)nmg due to dry sliding friction. Show that the oscillations are
isochronous (period independent of amplitude) with the amplitude of oscillation
decreasing by 2v.qlu% during each half-cycle until the mass comes to a stop.
[Hint: Use the result of Problem 33. When the force has a different algebraic
form at different times during the motion, as here, where the sign of the damping
force must be chosen so that the force is always opposed to the velocity, it is
necessary to solve the equation of motion separately for each interval of time
during which a particular expression for the force is to be used, and to choose
as initial conditions for each time interval the final position and velocity of the
preceding time interval.]
35. An undamped harmonic oscillator (7 = 0), initially at rest, is subject to
a force given by Eq. (2-191). (a) Find x(f). (b) For a fixed p Q , for what value
of dt is the final amplitude of oscillation greatest? (c) Show that as St -» 0, your
solution approaches that given by Eq, (2-190).
36. Find the solution analogous to Eq. (2-190) for a critically damped har-
monic oscillator subject to an impulse po delivered at t = to.
37. (a) Find, using the principle of superposition, the motion of an under-
damped oscillator [7 = (l/3)co ] initially at rest and subject, after t = 0,
to a force
F = A sin w t + B sin 3wot,
where wo is the natural frequency of the oscillator, (b) What ratio of B to A is
required in order for the forced oscillation at frequency 3wo to have the same
amplitude as that at frequency too?
38. Find, by the Fourier-series method, the steady-state solution for the
damped harmonic oscillator subject to a force
F(t) = I 0, if nT < '- ( n +*) r >
\f , if (n + %)T < t < (n -
(n + i)T < t < (n + 1)T,
where n is any integer, and T - 6jr/«o, where w is the resonance frequency of the
oscillator. Show that if 7 <JC coo, the motion is nearly sinusoidal with period T/3.
39. An underdamped oscillator initially at rest is acted upon, beginning at
t = 0, by a force
F = Foe-"'.
Find its motion by using Green's solution (2-210).
40. Using the result of Problem 36, find by Green's method the motion of a
critically damped oscillator initially at rest and subject to a force F(t).
CHAPTER 3
MOTION OF A PARTICLE IN TWO OR THREE DIMENSIONS
3-1 Vector algebra. The discussion of motion in two or three dimensions
is vastly simplified by the introduction of the concept of a vector. A
vector is denned geometrically as a physical quantity characterized by a
magnitude and a direction in space. Examples are velocity, force, and
position with respect to a fixed origin. Schematically, we represent a
vector by an arrow whose length and direction represent the magnitude
and direction of the vector. We shall represent a vector by a letter in bold-
face type. The same letter in ordinary italics will represent the magnitude
of the vector. (See Fig. 3-1.) The magnitude of a vector may also be
represented by vertical bars enclosing the vector symbol:
IAI.
(3-1)
Two vectors are equal if they have the same magnitude and direction;
the concept of vector itself makes no reference to any particular location.*
Fig. 3-1. A vector A and its magnitude A. (c > 0)
Fig. 3-2. Definition of multiplication of a vector by a scalar, (c > 0)
* A distinction is sometimes made between "free" vectors, which have no par-
ticular location in space; "sliding" vectors, which may be located anywhere along
a line; and "fixed" vectors, which must be located at a definite point in space. We
prefer here to regard the vector as distinguished by its magnitude and direction
alone, so that two vectors may be regarded as equal if they have the same magni-
tudes and directions, regardless of their positions in space.
68
3-1] VECTOR ALGEBRA 69
A quantity represented by an ordinary (positive or negative) number is
often called a scalar, to distinguish it from a vector. We define a product
of a vector A and a positive scalar c as a vector cA in the same direction as
A of magnitude cA. If c is negative, we define cA as having the magnitude
\c\A and a direction opposite to A. (See Fig. 3-2.) It follows from this
definition that
|cA| = |c| |A|. (3-2)
It is also readily shown, on the basis of this definition, that multiplication
by a scalar is associative in the following sense:
(cd)A = c(dA). (3-3)
It is sometimes convenient to be able to write the scalar to the right of the
vector, and we define Ac as meaning the same vector as cA:
Ac = cA. (3-4)
We define the sum (A + B) of two vectors A and B as the vector which
extends from the tail of A to the tip of B when A is drawn with its tip
at the tail of B, as in Fig. 3-3. This definition is equivalent to the usual
parallelogram rule, and is more convenient to use. It is readily extended
to the sum of any number of vectors, as in Fig. 3—4.
On the basis of the definition given in Fig. 3-3, we can readily prove
that vector addition is commutative and associative:
A + B = B + A, (3-5)
(A + B) + C=A + (B +.C). (3-6)
According to Eq. (3-6), we may omit parentheses in writing a vector sum,
since the order of adding does not matter. From the definitions given by
Figs. 3-2 and 3-3, we can also prove the following distributive laws:
c(A + B) = cA + cB, (3-7)
(c + d)A = cA + dk. (3-8)
These statements can be proved by drawing diagrams representing the
A + B
A+B+C+D
Fig. 3-3. Definition of addition of Fig. 3-4. Addition of several vec-
two vectors. tors.
70 MOTION OF PARTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3
(A + B) + C or A + (B + C)
Fig. 3-5. Proof of Eq. (3-6).
right and left members of each equation according to the definitions given.
For example, the diagram in Fig. 3-5 makes it evident that the result of
adding C to (A + B) is the same as the result of adding (B + C) to A.
According to Eqs. (3-3) through (3-8), the sum and product we have
defined have most of the algebraic properties of sums and products of
ordinary numbers. This is the justification for calling them sums and
products. Thus it is unnecessary to commit these results to memory.
We need only remember that we can manipulate these sums and products
just as we manipulate numbers in ordinary algebra with the one exception
that the product defined by Fig. 3-2 can be formed only between a scalar
and a vector, and the result is a vector.
A vector may be represented algebraically in terms of its components or
projections along a set of coordinate axes. Drop perpendiculars from the
tail and tip of the vector onto the coordinate axes as in Fig. 3-6. Then
the component of the vector along any axis is defined as the length of the
segment cut off on the axis by these perpendiculars. The component is
taken as positive or negative according to whether the projection of the
tip of the vector lies in the positive or negative direction along the axis
(a)
Fig. 3-6. (a) Components of a vector in a plane, (b) Components of a vector
in space.
3-1]
VECTOR ALGEBRA
71
Fig. 3-7. Diagrammatic proof of the formula A = Aj + A v j.
from the projection of the tail. The components of a vector A along x-, y-,
and z-axes will be written A x , A y , and A z . The notation (A x , A y , A z ) will
sometimes be used to represent the vector A:
A \A X , Ay, Az).
(3-9)
If we define vectors i, j, k of unit length along the x-, y-, z-axes respectively,
then we can write any vector as a sum of products of its components
with i, j, k:
A = A x \ + A y ] + A z k. (3-10)
The correctness of this formula can be made evident by drawing a dia-
gram in which the three vectors on the right, which are parallel to the
three axes, are added to give A. Figure 3-7 shows this construction for
the two-dimensional case.
We now have two equivalent ways of denning a vector: geometrically
as quantity with a magnitude and direction in space, or algebraically as a
set of three numbers [A x , A y , A z ), which we call its components.* The
operations of addition and multiplication by a scalar, which are defined
geometrically in Figs. 3-2 and 3-3 in terms of the lengths and directions of
the vectors involved, can also be defined algebraically as operations on the
components of the vectors. Thus cA is the vector whose components are
the components of A, each multiplied by c:
cA = (cA x , cA V) cA z ),
(3-11)
* These two ways of defining a vector are not quite equivalent as given here,
for the algebraic definition requires that a coordinate system be set up, whereas
the geometric definition does not refer to any particular set of axes. This flaw
can be remedied by making the algebraic definition also independent of any
particular set of axes. This is done by studying how the components change
when the axes are changed, and defining a vector algebraically as a set of three
quantities which transform in a certain way when the axes are changed. This
refinement will not concern us in this chapter.
72
MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS
y
(A + B),
[chap. 3
Fig. 3-8. Proof of equivalence of algebraic and geometric definitions of
vector addition.
and A + B is the vector whose components are obtained by adding the
components of A and B:
A + B = (A x + B x , A y + B V) A, + B z ).
(3-12)
The equivalence of the definitions (3-11) and (3-12) to the corresponding
geometrical definitions can be demonstrated by drawing suitable diagrams.
Figure 3-8 constitutes a proof of Eq. (3-12) for the two-dimensional case.
All vectors are drawn in Fig. 3-8 so that their components are positive;
for a complete proof, similar diagrams should be drawn for the cases where
one or both components of either vector are negative. The length of
a vector can be defined algebraically as follows:
|A| = (A* + AI+ At)
2-.1/2
(3-13)
where the positive square root is to be taken.
We can now give algebraic proofs of Eqs. (3-2), (3-3), (3-5), (3-6),
(3-7), and (3-8), based on the definitions (3-11), (3-12), and (3-13).
For example, to prove Eq. (3-7), we show that each component of the left
side agrees with each component on the right. For the a:-component, the
proof runs:
[c(A + B)L = c(A + B),
= c{A x + B x )
= cA x + cB x
= (cA), + (cB),
= (cA + cB) x .
[by Eq. (3-11)]
[by Eq. (3-12)]
[by Eq. (3-11)]
[by Eq. (3-12)]
3-1] VECTOR ALGEBRA 73
A-B>
B " 5"
Fig. 3-9. Two methods of subtraction of vectors.
Since all components are treated alike in the definitions (3-11), (3-12),
(3-13), the same proof holds for the y- and z-components, and hence the
vectors on the left and right sides of Eq. (3-7) are equal.
In view of the equivalence of the geometrical and algebraic definitions
of the vector operations, it is unnecessary, for geometrical applications, to
give both an algebraic and a geometric proof of each formula of vector
algebra. Either a geometric or an algebraic proof, whichever is easiest,
will suffice. However, there are important cases in physics where we have
to consider sets of quantities which behave algebraically like the com-
ponents of vectors although they cannot be interpreted geometrically as
quantities with a magnitude and direction in ordinary space. In order
that we may apply the rules of vector algebra in such applications, it is
important to know that all of these rules can be proved purely algebraically
from the algebraic definitions of the vector operations. The geometric
approach has the advantage of enabling us to visualize the meanings of
the various vector notations and formulas. The algebraic approach sim-
plifies certain proofs, and has the further advantage that it makes possible
wide applications of the mathematical concept of vector, including many
cases where the ordinary geometric meaning is no longer retained.
We may define subtraction of vectors in terms of addition and multi-
plication by —1:
A - B = A + (-B) = (A x - B x , A y - B y , A, - B z ). (3-14)
The difference A — B may be found geometrically according to either of
the two schemes shown in Fig. 3-9. Subtraction of vectors may be shown
to have all the algebraic properties to be expected by analogy with sub-
traction of numbers.
It is useful to define a scalar product (A-B) of two vectors A and B as
the product of their magnitudes times the cosine of the angle between them
(Fig. 3-10):
A-B = AB cos 6. (3-15)
The scalar product is a scalar or number. It is also called the dot product
or inner product, and can also be defined as the product of the magnitude of
either vector times the projection of the other along it. An example of
B.
74 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
its use is the expression for the work done when a force F acts through a
distance s not necessarily parallel to it:
W = Fs cosd = F-s.
We are entitled to call A-Ba product because it has the following alge-
braic properties which are easily proved from the geometrical definition
(3-15):
(cA)-B = A-(cB) = c(A-B), (3-16)
A-(B + C) = A-B + AC, (3-17)
A-B = B-A, (3-18)
Fig. 3-10. Angle between
A-A = A 2 (3-19) two vectors.
These equations mean that we can treat the dot product algebraically like
a product in the algebra of ordinary numbers, provided we keep in mind
that the two factors must be vectors and the resulting product is a scalar.
The following statements are also consequences of the definition (3-15),
where i, j, and k are the unit vectors along the three coordinate axes:
i-i = j-j = k-k = 1,
i-j = j-k = k-i = 0.
A-B = AB, when A is parallel to B, (3-21)
A-B = 0, when A is perpendicular to B. (3-22)
Notice that, according to Eq. (3-22), the dot product of two vectors is
zero if they are perpendicular, even though neither vector is of zero length.
The dot product can also be defined algebraically in terms of components:
A-B = A X B X + A y B y + A Z B Z . (3-23)
To prove that Eq. (3-23) is equivalent to the geometric definition (3-15),
we write A and B in the form given by Eq. (3-10), and make use of Eqs.
(3-16), (3-17), (3-18), and (3-20), which follow from Eq. (3-15):
A-B = (iA x + \A V + kA z )-(iB x + jB y + kB 2 )
= (i-i)A x B x + (i-j)A x B y + (i-k)4A + (j-i)A y B x + yjA y B y
+ ybA y B, + k-iA l B x + k-jA z B v + k-kA 2 B z
= A X B X + AyB y + A Z B Z .
This proves Eq. (3-23). The properties (3-16) to (3-20) can all be proved
(3-20)
3-1] VECTOR ALGEBRA 75
AxB
shaded area = |A x B|
B-—
A
Fig. 3-11. Definition of vector product.
readily from the algebraic definition (3-23) as well as from the geometric
definition (3-15). We can regard Eqs. (3-21) and (3-22) as algebraic
definitions of parallel and perpendicular.
Another product convenient to define is the vector product, also called
the cross product or outer product. The cross product (A X B) of two
vectors A and B is defined as a vector perpendicular to the plane of A and
B whose magnitude is the area of the parallelogram having A and B as
sides. The sense or direction of (A X B) is defined as the direction of
advance of a right-hand screw rotated from A toward B. (See Fig. 3-11.)
The length of (A X B), in terms of the angle between the two vectors,
is given by
| A X B| = AB sin 0. (3-24)
Note that the scalar product of two vectors is a scalar or number, while
the vector product is a new vector. The vector product has the following
algebraic properties which can be proved from the definition given in
Fig. 3-11:*
A X B = — B x A, (3-25)
(cA) X B = A X (cB) = c(A X B), (3-26)
A x (B + C) = (A x B) + (A x C), (3-27)
A x A = 0, (3-28)
A X B = 0, when A is parallel to B, (3-29)
|A X B| = AB, when A is perpendicular to B, (3-30)
ixi = jxj = kxk = 0,
i X j = k, j x k = i, k X i = j. (3-31)
* Here stands for the vector of zero length, sometimes called the null vector.
It has no particular direction in space. It has the properties:
A + = A, A-0 = 0, A X = 0, A — A = 0, = (0, 0, 0).
76 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
Hence the cross product can be treated algebraically like an ordinary-
product with the exception that the order of multiplication must not be
changed, and provided we keep in mind that the two factors must be vectors
and the result is a vector. Switching the order of factors in a cross product
changes the sign. This is the first unexpected deviation of the rules of
vector algebra from those of ordinary algebra. The reader should there-
fore memorize Eq. (3-25). Equations (3-29) and (3-30), as well as the
analogous Eqs. (3-21) and (3-22), are also worth remembering. (It goes
without saying that all geometrical and algebraic definitions should be
memorized.) In a repeated vector product like (A X B) X (C X D), the
parentheses cannot be omitted or rearranged, for the result of carrying out
the multiplications in a different order is not, in general, the same. [See,
for example, Eqs. (3-35) and (3-36).] Notice that according to Eq. (3-29)
the cross product of two vectors may be null without either vector being
the null vector.
From Eqs. (3-25) to (3-31), using Eq. (3-10) to represent A and B, we
can prove that the geometric definition (Fig. 3-11) is equivalent to the
following algebraic definition of the cross product:
A X B = (AyB, - A l B y , A,B X - A X B Z , A x B y - A y B x ). (3-32)
We can also write A X B as a determinant:
A x B =
i J k
A x Ay A z
B x By B z
(3-33)
Expansion of the right side of Eq. (3-33) according to the ordinary rules
for determinants yields Eq. (3-32). Again the properties (3-25) to (3-31)
follow also from the algebraic definition (3-32).
The following useful identities can be proved:
A-(B X C) = (A X B)C, (3-34)
A X (B X C) = B(A-C) - C(A-B), (3-35)
(A x B) x C = B(A-C) - A(B-C), (3-36)
i-(j X k) = 1. (3-37)
The first three of these should be committed to memory. Equation (3-34)
allows us to interchange dot and cross in the scalar triple product. The
quantity A-(B x C) can be shown to be the volume of the parallelepiped
whose edges are A, B, C, with positive or negative sign depending on
whether A, B, C are in the same relative orientation as i, j, k, that is,
depending on whether a right-hand screw rotated from A toward B would
3-2] APPLICATIONS TO A SET OF FORCES ACTING ON A PARTICLE 77
advance along C in the positive or negative direction. The triple vector
product formulas (3-35) and (3-36) are easy to remember if we note that
the positive term on the right in each case is the middle vector (B) times
the scalar product (A-C) of the other two, while the negative term is the
other vector within the parentheses times the scalar product of the other
two.
As an example of the use of the vector product, the rule for the force
exerted by a magnetic field of induction B on a moving electric charge q
(esu) can be expressed as
F = 2 V x B,
c
where c is the speed of light and v is the velocity of the charge. This
equation gives correctly both the magnitude and direction of the force.
The reader will remember that the subject of electricity and magnetism is
full of right- and left-hand rules. Vector quantities whose directions are
determined by right- or left-hand rules generally turn out to be expressible
as cross products.
3-2 Applications to a set of forces acting on a particle. According to the
principles set down in Section 1-3, if a set of forces Fi, F 2 , . . . , F„ act on a
particle, the total force F, which determines its acceleration, is to be ob-
tained by taking the vector sum of the forces F 1; F 2 , . . . , F„:
F = F x + F 2 H + F„. (3-38)
The forces F 1; F 2 , . . . , F„ are often referred to as component forces, and F
is called their resultant. The term component is here used in a more gen-
eral sense than in the preceding section, where the components of a vector
were defined as the projections of the vector on a set of coordinate axes.
When component is meant in this sense as one of a set of vectors whose sum
is F, we shall use the term (vector) component. In general, unless other-
wise indicated, the term component of a vector F in a certain direction will
mean the perpendicular projection of the vector F on a line in that direc-
tion. In symbols, the component of F in the direction of the unit vector n
is
F n = n-F. (3-39)
In this sense, the component of F is not a vector, but a number. The com-
ponents of F along the x-, y-, and 2-axes are the components in the sense of
Eq. (3-39) in the directions i, j, and k.
If the forces Fi, F 2 , . . . , F n are given, the sum may be determined
graphically by drawing a careful scale diagram according to the definition
of Fig. 3-3 or 3-4. The sum may also be determined analytically by
78 MOTION OF PARTICLE IN TWO OK THREE DIMENSIONS [CHAP. 3
F 2
Fig. 3-12. Sum of two forces.
drawing a rough sketch of the sum diagram and using trigonometry to
calculate the magnitude and direction of the vector F. If, for example,
two vectors are to be added, the sum can be found by using the cosine and
sine laws. In Fig. 3-12, F u F 2 , and 6 are given, and the magnitude and
direction of the sum F are calculated from
F 2 = F\ + F% - 2F X F 2 cos 6, (3-40)
Fi F* F
1 - 2 - (3-41)
sin $ sin a sin 9
Note that the first of these equations can be obtained by squaring, in the
sense of the dot product, the equation
F = Fj + F 2 . (3-42)
Taking the dot product of each member of this equation with itself, we
obtain
F-F = F 2 = Fi-F x + 2F!-F 2 + F 2 -F 2
= F\ + Fl — 2F X F 2 cos 0.
(Note that in Fig. 3-12 is the supplement of the angle between ¥ x and F 2
as defined by Fig. 3-10.) This technique can be applied to obtain directly
the magnitude of the sum of any number of vectors in terms of their
lengths and the angles between them. Simply square Eq. (3-38), and
split up the right side according to the laws of vector algebra into a sum of
squares and dot products of the component forces. The angle between F
and any of the component forces can be found by crossing or dotting the
component vector into Eq. (3-38). For example, in the case of a sum of
two forces, we cross ¥ x into Eq. (3^12) :
FiXF = FiXF, + FiX F 2 .
We take the magnitude of each side, using Eqs. (3-28) and (3-24):
Frf sin a = F X F 2 sin 8, or -!—- = J^- ■
sin 6 sin a
3-2] APPLICATIONS TO A SET OF FORCES ACTING ON A PARTICLE 79
When a sum of more than two vectors is involved, it is usually simpler to
take the dot product of the component vector with each side of Eq. (3-38).
The vector sum in Eq. (3-38) can also be obtained by adding separately
the components of Fi, . . . , F„ along any convenient set of axes:
Fx = Fix + F 2x + • • • + F nx ,
F y = F ly + F 2y + ■ ■ . + F nv ,
F z = F u + F 2l + • • • + F nl .
(3-43)
When a sum of a large number of vectors is to be found, this is likely to be
the quickest method. The reader should use his ingenuity in combining
and modifying these methods to suit the problem at hand. Obviously, if
a set of vectors is to be added which contains a group of parallel vectors,
it will be simpler to add these parallel vectors first before trying to apply
the methods of the preceding paragraph.
Fig. 3-13. Force F acting at point P.
Just as the various forces acting on a particle are to be added vectorially
to give the total force, so, conversely, the total force, or any individual
force, acting on a particle may be resolved in any convenient manner into
a sum of (vector) component forces which may be considered as acting
individually on the particle. Thus in the problem discussed in Section 1-7
(Fig. 1-4), the reaction force F exerted by the plane on the brick is resolved
into a normal component N and a frictional component /. The effect of
the force F on the motion of the brick is the same as that of the forces N
and / acting together. If it is desired to resolve a force F into a sum of
(vector) component forces in two or three perpendicular directions, this
can be done by taking the perpendicular projections of F in these directions,
as in Fig. 3-6. The magnitudes of the vector components of F, along a set
of perpendicular directions, are just the ordinary components of F in these
directions in the sense of Eq. (3-39).
If a force F in the a^-plane acts on a particle at the point P, we define
the torque, or moment of the force F about the origin (Fig. 3-13) as the
product of the distance OP and the component of F perpendicular to r:
No = rF sin a.
(3-44)
80
MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
The moment No of the force F about the point is denned as positive when
F acts in a counterclockwise direction about as in Fig. 3-13, and nega-
tive when F acts in a clockwise direction. We can define in a similar way
the moment about of any vector quantity located at the point P. The
concept of moment will be found useful in our study of the mechanics of
particles and rigid bodies. The geometrical and algebraic properties of
torques will be studied in detail in Chapter 5. Notice that torque can be
denned in terms of the vector product:
No = ±|r X F|.
(3-45)
where the + or — sign is used according to whether the vector r X F points
in the positive or negative direction along the z-axis.
Fig. 3-14. Moment of a force about an axis in space.
We can generalize the above definition of torque to the three-dimensional
case by denning the torque or moment of a force F, acting at a point P,
about an axis AB (Fig. 3-14). Let n be a unit vector in the direction of
AB, a nd l et F be resolved into vector components parallel and perpendicu-
lar to AB:
F = F|| + F±, (»-46)
where
Fii = n(n-F),
(3-47)
Fj. = F - Fn. .
We now define the moment of F about the axis AB as the moment, denned
by Eq. (3-44) or (3-45), of the force F±, in a plane through the point P
3-3] DIFFERENTIATION AND INTEGRATION OF VECTORS 81
perpendicular to AB, about the point at which the axis ~AB passes through
this plane:
Nab = dbrFxsina = ±|r X F x |, (3-48)
where the + or — sign is used, depending on whether r x Fx is in the same
or opposite direction to n. According to this definition, a force like Fn
parallel to AB has no torque or moment about A~B. Since r X Fn is per-
pendicular to n,
n-(r x F) = n-[r x (F M + Fx)] #
= n-(r x Fn) +n-(r x Fx)
= n-(r x F±)
= ±|r x F x |.
Hence we can define Nab in a neater way as follows:
N AB = n-(r X F). (3-49)
This definition automatically includes the proper sign, and does not require
a resolution of F into Fn and Fx- Furthermore, r can now be drawn to P
from any point on the axis AB, since a component of r parallel to A~B, like
a component of F parallel to AB, gives a component in the cross product
perpendicular to n which disappears from the dot product.
Equation (3-49) suggests the definition of a vector torque or vector moment,
about a point 0, of a force F acting at a point P, as follows:
N = r X F, (3-50)
where r is the vector from to P. The vector torque N has, according
to Eq. (3-49), the property that its component in any direction is the
torque, in the previous sense, of the force F about an axis through in that
direction. Hereafter the term torque will usually mean the vector torque
defined by Eq. (3-50). Torque about an axis AB in the previous sense
will be called the component of torque along ~AB. We can define the
vector moment of any vector located at a point P, about a point 0, by an
equation analogous to Eq. (3-50).
3-3 Differentiation and integration of vectors. A vector A may be a
function of a scalar quantity, say t, in the sense that with each value of t a
certain vector A(t) is associated, or algebraically in the sense that its com-
ponents may be functions of t:
A = A(<) = [A x (t), A y {t), AM- (3-51)
82 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
The most common example is that of a vector function of the time; for
example, the velocity of a moving particle is a function of the time: v(t).
Other cases also occur, however; for example, in Eq. (3-76), the vector n
is a function of the angle 6. We may define the derivative of the vector A
with respect to t in analogy with the usual definition of the derivative of
a scalar function (see Fig. 3-15) :
§ = lim A « + *> ~ A ® • (3-52)
dt a«-»o At
(Division by At here means multiplication by 1/At.) We may also define
the vector derivative algebraically in terms of its components:
dA
dt
(dA x dA y dA z \ . dA x , ,dA y . ,dA z ., eo .
As an example, if v(<) is the vector velocity of a particle, its vector accelera-
tion a is
a = dv/dl.
Examples of the calculation of vector derivatives based on either definition
(3-52) or (3-53) will be given in Sections 3-4 and 3-5.
The following properties of vector differentiation can be proved by
straightforward calculation from the algebraic definition (3-53), or they
may be proved from the definition (3-52) in the same way the analogous
properties are proved for differentiation of a scalar function:
s< A + B > = f + f' ™
£(/A)_fA+/f, (3-55)
|(A.B, = f.B + A.f, (3-56,
|(AxB) = f XB + Axf- (3-57)
These results imply that differentiation of vector sums and products obeys
the same algebraic rules as differentiation of sums and products in ordinary
calculus, except, however, that the order of factors in the cross product
must not be changed [Eq. (3-57)]. To prove Eq. (3-55), for example,
from the definition (3-53), we simply show by direct calculation that the
corresponding components on both sides- of the equation are equal, making
use of the definitions and properties of the vector operations introduced
3-3] DIFFERENTIATION AND INTEGRATION OF VECTORS 83
in the preceding section. For the z-component, the proof runs:
d
[>>!=
dt
(/A), [by Eq. (3-53)]
= J t CM.) [by Eq. (3-11)]
= it a 4. f dAx [standard rule of ordi-
dt x dt nary calculus]
dt A * + Kw) x ^y Eq. (3-53)]
dt J
Xdi A ) x + \ f -d7) x
(df dA\
\dt A+J dt) x
[by Eq. (3-11)]
[by Eq. (3-12)]
As another example, to prove Eq. (3-56) from the definition (3-52), we
proceed as in the proof of the corresponding theorem for products of ordi-
nary scalar functions. We shall use the symbol A to stand for the increment
in the values of any function between t and t + At; the increment AA of a
vector A is defined in Fig. 3-15. Using this definition of A, and the rules
of vector algebra given in the preceding section, we have
A(A-B) = (A + AA).(B + AB) - A-B
At At
= (AA)-B + A-(AB) + (AA)-(AB)
At
(AA)-B A-(AB) (AA)-(AB)
At ^ At + At
AA AB (AA)-(AB)
At ^ At ^ At
(3-58)
When At — » 0, the left side of Eq. (3-58) approaches the left side of Eq.
(3-56), and the first two terms on the right side of Eq. (3-58) approach the
A(t + At)
Fig. 3-15. Vector increment AA = A(t + At) — A(t).
84 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
Fig. 3-16. The position vector r of the point (x, y, z).
two terms on the right of Eq. (3-56), while the last term on the right of
Eq. (3-58) vanishes. The rigorous justification of this limit process is
exactly similar to the justification required for the corresponding process
in ordinary calculus.
In treating motions in three-dimensional space, we often meet scalar
and vector quantities which have a definite value at every point in space.
Such quantities are functions of the space coordinates, commonly x, y,
and z. They may also be thought of as functions of the position vector r
from the origin to the point x, y, z (Fig. 3-16). We thus distinguish scalar
point functions
«(r) = u{x, y, z),
and vector point functions
A(r) = A(x, y, z) = [A x (x, y, z), A y (x, y, z), A z (x, y, z)].
An example of a scalar point function is the potential energy V(x, y, z) of a
particle moving in three dimensions. An example of a vector point func-
tion is the electric field intensity E(x, y, z). Scalar and vector point func-
tions are often functions of the time t as well as of the point x, y, z in space.
If we are given a curve C in space, and a vector function A defined at
points along this curve, we may consider the line integral of A along C:
f A-dr.
Jc
To define the line integral, imagine the curve C divided into small seg-
ments, and let any segment be represented by a vector dr in the direction
of the segment and of length equal to the length of the segment. Then
the curve consists of the successive vectors dr laid end to end. Now for
each segment, form the product A-dr, where A is the value of the vector
function at the position of that segment. The line integral above is denned
as the limit of the sums of the products A-dr as the number of segments
3-3] DIFFERENTIATION AND INTEGRATION OF VECTORS 85
increases without limit, while the length |dr| of every segment approaches
zero. As an example, the work done by a force F, which may vary from
point to point, on a particle which moves along a curve C is
W
= [ F-dr,
Jc
which is a generalization, to the case of a varying force and an arbitrary
curve C, of the formula
W = F-s,
for a constant force acting on a body moving along a straight line seg-
ment s. The reason for using the symbol dx to represent a segment of the
curve is that if r is the position vector from the origin to a point on the
curve, then dx is the increment in r (see Fig. 3-15) from one end to the
other of the corresponding segment. If we write r in the form
r = ix + \y + kz, (3-59)
then
dx = i dx + j dy + k dz, (3-60)
where dx, dy, dz are the differences in the coordinates of the two ends of the
segment. If s is the distance measured along the curve from some fixed
point, we may express the line integral as an ordinary integral over the
coordinate s:
f A-dx = f A cos dds, (3-61)
where 6 is the angle between A and the tangent to the curve at each point.
(See Fig. 3-17.) This formula may be used to evaluate the integral if
we know A and cos 6 as functions of s. We may also write the integral,
Fig. 3-17. Elements involved in the line integral.
86
MOTION OF PARTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3
using Eq. (3-60), as
f A'dt = f (A x dx + Aydy + A. z dz).
Jc Jc
(3-62)
One of the most convenient ways to represent a curve in space is to give
the three coordinates (x, y, z) or, equivalently, the position vector r, as
functions of a parameter s which has a definite value assigned to each
point of the curve. The parameter s is often, though not necessarily, the
distance measured along the curve from some reference point, as in Fig.
3-17 and in Eq. (3-61). The parameter s may also be the time at which
a moving particle arrives at any given point on the curve. If we know
A(r) and r(s), then the line integral can be evaluated from the formula
//■*- /(*•§)*
/(^5 + ^ + ^S>
(3-63)
The right member of this equation is an ordinary integral over the vari-
able s.
Figure 3-18
As an example of the calculation of a line integral, let us compute the
work done on a particle moving in a semicircle of radius a about the origin
in the a^-plane, by a force attracting the particle toward the point (a; = a,
y = 0) and proportional to the distance of the particle from the point
(a, 0). Using the notation indicated in Fig. 3-18, we can write down the
following relations:
= iOr - a),
7T
2
e = ^ - P = ha,
D 2 = 2a 2 (l - cos a),
D — 2a sin •
F = -fcD,
JcD = 2ka sin -^ >
s = o(ir — a).
3-4] KINEMATICS IN A PLANE 87
Using these relations, we can evaluate the work done, using Eq. (3-61):
W = f F>dt
Jc
(■xa
= F cosd ds
Js=0
= — J 2ka 2 sin ^ cos „ da
= — 4fca 2 f sin cos dO
2ka
2
In order to calculate the same integral from Eq. (3-63), we express r and
F along the curve as functions of the parameter a:
x = a cos a, y = a sin a,
2 a
F x = kD cos /S = 2ka sin 2 | = fca(l — cos a),
f„ = — kD sin /3 = — 2ka sin - cos ^ = — &a sin a.
The work is now, according to Eq. (3-63),
W = f F-dr
-/!('•£+'.£)*"
/•o
= / [— fca 2 (l — cos a) sin a — ka 2 sin a cos a] da
J V
= fca 2 / sin a da
Jo
= 2fta 2 .
3-4 Kinematics in a plane. Kinematics is the science which describes
the possible motions of mechanical systems without regard to the dynami-
cal laws that determine which motions actually occur. In studying the
kinematics of a particle in a plane, we shall be concerned with methods for
describing the position of a particle, and the path followed by the particle,
and with methods for rinding the various components of its velocity and
acceleration.
88 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
Fig. 3-19. Position vector and rectangular coordinates of a point P in a plane.
The simplest method of locating a particle in a plane is to set up two
perpendicular axes and to specify any position by its rectangular coordi-
nates x, y with respect to these axes (Fig. 3-19). Equivalently, we may
specify the position vector r = (x, y) from the origin to the position of the
particle. If we locate a position by specifying the vector r, then we need
to specify in addition only the origin from which the vector is drawn.
If we specify the coordinates x, y, then we must also specify the coordi-
nate axes from which x, y are measured.
Having set up a coordinate system, we next wish to describe the path
of a particle in the plane. A curve in the xy-plane may be specified by
giving y as a function of x along the curve, or vice versa:
or
y = v(x),
x = x(y).
(3-64)
(3-65)
Forms (3-64) and (3-65), however, are not convenient in many cases, for
example when the curve doubles back on itself. We may also specify the
curve by giving a relation between x and y,
/(*, y) = 0,
(3-66)
such that the curve consists of those points whose coordinates satisfy this
relation. An example is the equation of a circle:
x 2 + y 2
,2 _
0.
One of the most convenient ways to represent a curve is in terms of a
parameter s:
x = x(s), y = y(s), (3-67)
or
r = r(s).
The parameter s has a unique value at each point of the curve. As s
varies, the point [x(s), y(s)] traces out the curve. The parameter s may,
3-4] KINEMATICS IN A PLANE 89
for example, be the distance measured along the curve from some fixed
point. The equations of a circle can be expressed in terms of a parameter 6
in the form
x = a cos 6,
y = a sin 6,
where 6 is the angle between the z-axis and the radius a to the point (x, y)
on the circle. In terms of the distance s measured around the circle,
s
x — a cos - >
a
. s
y = a sin - •
a
In mechanical problems, the parameter is usually the time, in which
case Eqs. (3-67) specify not only the path of the particle, but also the
rate at which the particle traverses the path. If a particle travels with
constant speed v around a circle, its position at any time t may be given by
vt
x = a cos — i
a
. vt
y = a sin — •
a
If a particle moves along the path given by Eq. (3-67), we may specify
its motion by giving s(i), or by specifying directly
x = x(t), y = y(t), (3-68)
or
r = t{t). (3-69)
The velocity and acceleration, and their components, are given by
v = — = i — -I- i ^
dt dt^~ J dt
dx dy
v *=di' v « = Tt'
dv _ d^r _ . d?x . . d?y
dt ~ dl* ~ l dP + J dt 2 '
— d?£ _ d 2 y
i* - dt 2 ' a y~dj2-
(3-70)
(3-71)
90 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
\*
\V^ n
T
y
_L
n(9 + dB)f\<fa
tL
Fig. 3-20. Plane polar coordinates.
Fig. 3-21. Increments in the vec-
tors n and 1.
Polar coordinates, shown in Fig. 3-20, are convenient in many prob-
lems. The coordinates r, are related to x, y by the following equations:
x = r cos i
y = r sin 0,
and
= tan x - = sin
x
r= (x< + y*)
-l y
(x 2 + ?/2)l/2
2x1/2
(3-72)
(3-73)
= cos
(* 2 + y 2 ) 1 ' 2
We define unit vectors n, 1 in the directions of increasing r and 0, respec-
tively, as shown. The vectors n, 1 are functions of the angle 6, and are
related to i, j by the equations
n = i cos 8 + j sin 0,
1 = — i sin 9 + j cos 6.
(3-74)
Equations (3-74) follow by inspection of Fig. 3-20. Differentiating, we
obtain the important formulas
eta _ . rfl _
dd ~ ' dd~
-n.
(3-75)
Formulas (3-75) can also be obtained by studying Fig. 3-21 (remembering
that |n| = |1| = 1). The position vector r is given very simply in terms
of polar coordinates:
r = rn(0). (3-76)
We may describe the motion of a particle in polar coordinates by specifying
r(t), 0(0, thus determining the position vector r(t). The velocity vector is
dr dr . da dO . . M
dt dt dd dt
(3-77)
3-5] KINEMATICS IN THREE DIMENSIONS 91
Thus we obtain the components of velocity in the n, 1 directions:
v r = f, v e = rd. (3-78)
The acceleration vector is
dt dO dt dO dt
= (f - r^ 2 )n + (WJ + 2rt)L (3-79)
The components of acceleration are
a r = r — r0 2 , a„ = r0 + 2^. (3-80)
The term rd 2 = v 2 /r is called the centripetal acceleration arising from motion
in the direction. If r = f = 0, the path is a circle, and a r = —v 2 /r.
This result is familiar from elementary physics. The term 2f6 is some-
times called the coriolis acceleration.
3-5 Kinematics in three dimensions. The development in the preceding
section for kinematics in two dimensions utilizing rectangular coordinates
can be extended immediately to the three-dimensional case. A point is
specified by its coordinates x, y, z, with respect to chosen rectangular axes
in space, or by its position vector r = (x, y, z) with respect to a chosen
origin. A path in space may be represented in the form of two equations
in x, y, and z:
f(x, y, z) = 0, g(x, y, z) = 0. (3-81)
Each equation represents a surface. The path is the intersection of the
two surfaces. A path may also be represented parametrically:
x = x(s), y = y(s), z = z(s). (3-82)
Velocity and acceleration are again given by
dr
and
v = ^i = w x + )v y + kv z ,
(3-83)
dx dy dz
v * = Tt' v « = Tt' v °=di'
(3-84)
dy . . . , ,
a = -^ = ia x + ja y + ka z ,
(3-85)
d 2 x d 2 y d 2 z
'■ ~ dp ' ay ~~dp' az ~ Ifi '
(3-86)
Many coordinate systems other than cartesian are useful for special
problems. Perhaps the most widely used are spherical polar coordinates
92 MOTION OF PARTICLE IN TWO OK THREE DIMENSIONS [CHAP. 3
<
1
j
c
A
i/
z
1 y^K
x/ \ p
/ V
Fig. 3-22. Cylindrical polar coordinates.
and cylindrical polar coordinates. Cylindrical polar coordinates (p, tp, z)
are denned as in Fig. 3-22, or by the equations
x — p cos tp, y = p sin tp, z = z,
and, conversely,
p= (x 2 + y 2 ) 112 ,
(3-87)
^ -ill • -l V
<P — tan - = sin , J
a; (x 2 -f- 2/2)1/2
2 = 2.
= cos
(x 2 + y 2 )i/ 2
(3-88)
A system of unit vectors h, m, k, in the directions of increasing p, tp, z, re-
spectively, is shown in Fig. 3-22. k is constant, but m and h are func-
tions of tp, just as in plane polar coordinates:
h = i cos <p + j sin tp, m = — i sin <p + j cos <p, (3-89)
and, likewise,
dh
dtp
— m,
dm
dtp
(3-90)
The position vector r can be expressed in cylindrical coordinates in the
form
t = ph + zk. (3-91)
Differentiating, we obtain for velocity and acceleration, using Eq. (3-90) :
v = ^ = ph + p4>m + zk, (3-92)
a = J = (p - P? 2 )h + (p£ + 2p£)m + zk. (3-93)
3-5]
KINEMATICS IN THREE DIMENSIONS
93
Since k, m, h form a set of mutually perpendicular unit vectors, any vector
A can be expressed in terms of its components along k, m, h:
A = Aph -f A^m + Ajs.
(3-94)
It must be noted that since h and m are functions of <p, the set of com-
ponents {A p , A v , A z ) refers in general to a specific point in space at which
the vector A is to be located, or at least to a specific value of the coordi-
nate <p. Thus the components of a vector in cylindrical coordinates, and
in fact in all systems of curvilinear coordinates, depend not only on the
vector itself, but also on its location in space. If A is a function of a
parameter, say t, then we may compute its derivative by differentiating
Eq. (3-94), but we must be careful to take account of the variation of h
and m if the location of the vector is also changing with t (e.g., if A is the
force acting on a moving particle) :
dk
dt
-te-^iM^+^s)
dA z .
m + ~dT k -
(3-95)
Formulas (3-92) and (3-93) are special cases of Eq. (3-95). A formula
for dA/dt could have been worked out also for the case of polar coordinates
in two dimensions considered in the preceding section, and would, in fact,
have been exactly analogous to Eq. (3-95) except that the last term would
be missing.
Spherical polar coordinates (r, 6, tp) are defined as in Fig. 3-23 or by the
equations
r sin 6 cos <p, y = r sin 8 sin <p, z = r cos i
(3-96)
The expressions for x and y follow if we note that p = r sin 6, and
Fig. 3-23. Spherical polar coordinates.
94 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
use Eq. (3-87) ; the formula for z is evident from the diagram. Conversely,
/ 2 1 2 r 2\l/2
r = {x' + y' + z Y ,
! (x 2 4- v 2 ) 112
= tan" 1 {X + g y ' , (3-97)
<p = tan -1 --
x
Unit vectors n, 1, m appropriate to spherical coordinates are indicated in
Fig. 3-23, where m is the same vector as in cylindrical coordinates. The
unit vector h is. also useful in obtaining relations involving n and 1. We
note that k, h, n, 1, all lie in one vertical plane. From the figure, and
Eq. (3-89), we have
n = k cos + h sin 8 = k cos + i sin cos <p + j sin sin <p,
1 = — k sin 8 + h cos 6 = — k sin 8 + i cos 8 cos <p + j cos 8 sin <p, (3-98)
m = — i sin tp -\- j cos <p.
By differentiating these formulas, or more easily by inspection of the dia-
gram (as in Fig. 3-21), noting that variation of 8, with <p and r fixed, corre-
sponds to rotation in the k, n, h, 1 plane, while variation of <p, with 8 and r
fixed, corresponds to rotation around the z-axis, we find
(3-99)
88
an . „
t- = m sm 8,
dip
dl
91
T- = m cos 8,
0<p
^=0
ae '
dm . . n
-r— = — h = — n sm 8
o<p
1 cos 8.
In spherical coordinates the position vector is simply
r = ra(0, <p). (3-100)
Differentiating and using Eqs. (3-99), we obtain the velocity and accelera-
tion:
v = -j t = fn + rfa + (/y sin 0)m, (3-101)
a = -^ = (r - r6 2 — rip 2 sin 2 0)n + (r9 + 2r6 - rip 2 sin 8 cos 0)1
+ (rv sin 8 + 2f<p sin + 2rd<p cos 0)m. (3-102)
3-6] ELEMENTS OF VECTOR ANALYSIS 95
Again, n, m, 1 form a set of mutually perpendicular unit vectors, and any
vector A may be represented in terms of its spherical components:
A = Ara + A e l + Ajn. (3-103)
Here again the components depend not only on A but also on its location.
If A is a function of t, then
dk _ f dA r
dt ~\dt
A e
dd A ■ a ^¥> 1
,(dA,
^Xdt
+ A 41-
+ Ar dt
-A,
, cos t
•*)'
,(dA<p
+ Vdt~
+ A r sin 6
d<p
dt
+ A e
-•S)
m. (3-104)
3-6 Elements of vector analysis. A scalar function u{x, y, z) has three
derivatives, which may be thought of as the components of a vector point
function called the gradient of u:
We may also define grad u geometrically as a vector whose direction is the
direction in which u increases most rapidly and whose magnitude is the
directional derivative of u, i.e., the rate of increase of u per unit distance,
in that direction. That this geometrical definition is equivalent to the
algebraic definition (3-105) can be seen by taking the differential of u:
du ^di dx + dy- dy + Tz dz - ( 3 - 106 )
Equation (3-106) has the form of a scalar product of grad u with the vector
dr whose components are dx, dy, dz:
du = dr-grad u. (3-107)
Geometrically, du is the change in u when we move from the point r =
(x, y, z) to a nearby point r + dt — {x + dx, y + dy, z + dz). By Eq.
(3-15):
du — \dt\ |grad u\ cos 6, (3-108)
where is the angle between dt and grad u. Thus at a fixed small dis-
tance \dx\ from the point r, the change in u is a maximum when dx is in the
same direction as grad u, and then:
|gradM| = -prT
96 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
This confirms the geometrical description of grad u given above. An al-
ternative geometrical definition of grad u is that it is a vector such that
the change in u, for an arbitrary small change of position dx, is given by
Eq. (3-107).
In a purely symbolic way, the right member of Eq. (3-105) can be
thought of as the "product" of a "vector":
_ ( d d d\ . d . . d , , d , n ,._.
V = U'^'^ = 1 ^ + , ^ + k ai' (3 - 109)
with the scalar function u:
gradw = Vm. (3-110)
The symbol V is pronounced "del." V itself is not a vector in the geo-
metrical sense, but an operation on a function u which gives a vector Vw.
However, algebraically, V has properties nearly identical with those of a
vector. The reason is that the differentiation symbols (d/dx, d/dy, d/dz)
have algebraic properties like those of ordinary numbers except when they
act on a product of functions:
and
d , , , du . dv d d d d . .
_(„ + „) = _ + _, __ M = __ W , (3-111)
f x (au) = a d £, (3-112)
provided a is constant. However,
d , n du dv . .
_(„)«_, + „__. (3-H3)
In this one respect differentiation operators differ algebraically from ordi-
nary numbers. If d/dx were a number, d/dx(uv) would equal either
u(d/dx)v or v{d/dx)u. Thus we may say that d/dx behaves algebraically
as a number except that when it operates on a product, the result is a sum
of terms in which each factor is differentiated separately, as in Eq. (3-113).
A similar remark applies to the symbol V. It behaves algebraically as a
vector, except that when it operates on a product it must be treated also
as a differentiation operation. This rule enables us to write down a large
number of identities involving the V symbol, based on vector identities.
We shall require very few of these in this text, and shall not list them here. *
* For a more complete treatment of vector analysis, see H. B. Phillips, Vector
Analysis. New York: John Wiley & Sons, 1933.
3-6] ELEMENTS OF VECTOR ANALYSIS 97
Fig. 3-24. A volume V bounded by a surface 5.
We can form the scalar product of V with a vector point function
A(x, y, z). This is called the divergence of A:
• A -* A -'£+%+'■&■ < 3 ->">
The geometrical meaning of div A is given by the following theorem, called
the divergence theorem, or Gauss' theorem:
jjj VA dV = Jfn>AdS, (3-115)
v s
where V is a given volume, S is the surface bounding the volume V, and n
is a unit vector perpendicular to the surface S pointing out from the volume
at each point of 5 (Fig. 3-24). Thus n-A is the component of A normal
to S, and Eq. (3-115) says that the "total amount of V-A inside V" is
equal to the "total flux of A outward through the surface 5." If v repre-
sents the velocity of a moving fluid at any point in space, then
//n.vdS
a
represents the volume of fluid flowing across S per second. If the fluid is
incompressible, then according to Eq. (3-115),
///
v-vdV
would represent the total volume of fluid being produced within the vol-
ume V per second. Hence v-v would be positive at sources from which
the fluid is flowing, and negative at "sinks" into which it is flowing. We
omit the proof of Gauss' theorem [Eq. (3-115)]; it may be found in any
book on vector analysis.*
Sec, e.g„ Phillips, op. cit. Chapter 3, Section 32.
98 MOTION OF PARTICLE IN' TWO OB THREE DIMENSIONS [CHAP. 3
Fig. 3-25. A surface S bounded by a curve C.
We can also form a cross product of V with a vector point function
A(x, y, z). This is called the curl of A:
The geometrical meaning of the curl is given by Stokes' theorem:
j jn-(v x A)dS = f A-dr, (3-117)
where S is any surface in space, n is the unit vector normal to S, and C is
the curve bounding S, dr being taken in that direction in which a man would
walk around C if his left hand were on the inside and his head in the direc-
tion of n. (See Fig. 3-25.) According to Eq. (3-117), curl A at any point
is a measure of the extent to which the vector function A circles around that
point. A good example is the magnetic field around a wire carrying an
electric current, where the curl of the magnetic field intensity is propor-
tional to the current density. We omit the proof of Stokes' theorem [Eq.
(3-117)].*
The reader should not be bothered by the difficulty of fixing these ideas
in his mind. Understanding of new mathematical concepts like these
comes to most people only slowly, as they are put to use. The definitions
are recorded here for future use, One cannot be expected to be familiar
with them until he has seen how they are used in physical problems.
The symbolic vector V can also be expressed in cylindrical coordinates in terms
of its components along h, m, k. (See Fig. 3-22.) We note that if u = u(p, tp, z),
* ->+£*+£*■ (3 - 118)
For the proof see Phillips, op. tit. Chapter 3, Section 29.
3-6] ELEMENTS OF VECTOR ANALYSIS 99
and, from Eqs. (3-91) and (3-90),
dr = h dp + rap d<p + k dz, (3-119)
a result whose geometric significance will be evident from Fig. 3-22. Hence, if we
write
_ . d . m d , , d
we will have, since h, m, k are a set of mutually perpendicular unit vectors,
du = dr-Vw, (3-121)
as required by the geometrical definition of Vu = grad u. [See the remarks
following Eq. (3-107).] A formula for V could have been worked out also for
the case of polar coordinates in two dimensions and would have been exactly
analogous to Eq. (3-120) except that the term in z would be missing. In apply-
ing the symbol V to expressions involving vectors expressed in cylindrical co-
ordinates [Eq. (3-94)], it must be remembered that the unit vectors h and m are
functions of <p and subject to differentiation when they occur after d/d<p.
We may also find the vector V in spherical coordinates (Pig. 3-23) by noting
that
, du , . du ,. , du ,
du = Tr dr + S6 M + dl P d *> ( 3 ~ 122 )
and
dt = n dr + Xr dO + mr sin d<p. (3-123)
Hence
_ 3,13. m d
V=n 3r + ^+rlu7fl^' ( 3 ~ 124 >
in order that Eq. (3-121) may hold. Again we caution that in working with Eq.
(3-124), th3 dependence of n, 1, m on 0, <p must be kept in mind. For example,
the divergence of a vector function A expressed in spherical coordinates [Eq.
(3-103)] is
VA = n-— • + -••— +
dr r dd r sin d<p
dA
dr
'■ +i M+ A )+7h,(%+ *-•"+ *-•)
= dAr 2Ar 1 dAt _JU_ , _1__ dA?
dr r ~*~ r dd r tan r sin dip
- 1 d /- 2 -^ i 1 d - • «^ , I dA„
-r-2fr (rA ' ) + 7^0r0 (sm6A ° ) + 7^re-dV-
(In the above calculation, we use the fact that 1, m, n are a set of mutually per-
pendicular unit vectors.)
100 MOTION OP PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
3-7 Momentum and energy theorems. Newton's second law, as formu-
lated in Chapter 1, leads, in two or three dimensions, to the vector equation
m§ = F. (3-125)
In two dimensions, this is equivalent to two component equations, in three
dimensions, to three, which are, in cartesian coordinates,
d 2 x „ d 2 y „ _ d 2 z
m -- = F x , m^i = F m m — = F z . (3-126)
In this section, we prove, using Eq. (3-125), some theorems for motion in
two or three dimensions which are the vector analogs to those proved in
Section 2-1 for one-dimensional motion.
The linear momentum vector p of a particle is to be defined, according
to Eq. (1-10), as follows:
p = mv. (3-127)
Equations (3-125) and (3-126) can then be written
i <»*> = I = f > (3 - i28)
or, in component form,
d ^ = F x , *g = F n f = *V (3-129)
If we multiply by dt, and integrate from t x to t 2 , we obtain the change in
momentum between t\ and t^.
p2 — Pi = «v 2 — ravi = J F dt. (3-130)
The integral on the right is the impulse delivered by the force, and is a
vector whose components are the corresponding integrals of the com-
ponents of F. In component form:
rh
Px 2 — p Xl = / Fx dt,
Jti
Vv 2 - Pvi = [* Fydt, (3-131)
Pz 2 — Pzi = f 2 p z dt.
In order to obtain an equation for the rate of change of kinetic energy,
we proceed as in Section 2-1, multiplying Eqs. (3-126) by v x , v y , v z , respec-
tively, to obtain
| (W) = F x v x , j t (inwl) = F y v y , j t (imv 2 ) = F z v z . (3-132)
3-8] PLANE AND VECTOR ANGULAR MOMENTUM THEOREMS 101
Adding these equations, we have
d
dt
or
- [*f»(»2 + V 2 y + Vf)\ = F X V X + FyVy + F z v z
d ,, 2^ dT
-(J TOy ) = W = F.v. (3-133)
This equation can also be deduced from the vector equation (3-125) by
taking the dot product with v on each side, and noting that
d , 2i d . . dv , dv „ dv
^(0=^(v.v) = -.v + v.^=2v.^.
Thus, by Eq. (3-132),
¥ ' v =™'Tt = * m dt=dt^ mv) -
Multiplying Eq. (3-133) by dt, and integrating, we obtain the integrated
form of the energy theorem:
T 2 - T x = \mo\ - \mv\ = pF-vett. (3-134)
Since v dt = dr, if F is given as a function of r, we can write the right
member of Eq. (3-134) as a line integral:
T a -T l = Pp-dr, (3-135)
Jri
where the integral is to be taken along the path followed by the particle
between the points i x and r 2 . The integral on the right in Eqs. (3-134) and
(3-135) is the work done on the particle by the force between the times
ti and t 2 - Note how the vector notation brings out the analogy between
the one- and the two- or three-dimensional cases of the momentum and
energy theorems.
3-8 Plane and vector angular momentum theorems. If a particle moves
in a plane, we define its angular momentum Lo about a point as the
moment of its momentum vector about the point 0, that is, as the product
of its distance from times the component of momentum perpendicular
to the line joining the particle to 0. The subscript will usually be omitted,
except when moments about more than one origin enter into the discussion,
but it must be remembered that angular momentum, like torque, refers
to a particular origin about which moments are taken. The angular
momentum L is taken as positive when the particle is moving in a counter-
clockwise sense with respect to 0; L is expressed most simply in terms of
polar coordinates with as origin. Let the particle have mass m. Then
102 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
Fig. 3-26. Components of velocity in a plane.
its momentum is mv, and the component of momentum perpendicular to
the radius vector from is mv t (Fig. 3-26), so that, if we use Eq. (3-78),
L = rmv e = mrH. (3-136)
If we write the force in terms of its polar components:
F = nF r + IF,, (3-137)
then in plane polar coordinates the equation of motion, Eq. (3-125), be-
comes, by Eq. (3-80),
ma r = mr — mrd 2 = F n (3-138)
ma, = mrS + 2mr& = F e . (3-139)
We now note that
= 2mrf6 + mr 2 8.
dt
Thus, multiplying Eq. (3-139) by r, we have
(mr 2 6) = rF e = N. (3-140)
dL d , 2/n
dt dt
The quantity rF e is the torque exerted by the force F about the point 0.
Integrating Eq. (3-140), we obtain the integrated form of the angular
momentum theorem for motion in a plane:
L 2 - L t = mr%6 2 - mr\»i = f" rF e dt. (3-141)
We can generalize the definition of angular momentum to apply to three-
dimensional motion by defining the angular momentum of a particle about
an axis in space as the moment of its momentum vector about this axis,
just as in Section 3-2 we denned the moment of a force about an axis.
3-8] PLANE AND VECTOR ANGULAR MOMENTUM THEOREMS
103
The development is most easily carried out in cylindrical coordinates with
the z-axis as the axis about which moments are to be taken. The generali-
zation of theorems (3-140) and (3-141) to this case is then easily proved in
analogy with the proof given above. This development is left as an exercise.
As a final generalization of the concept of angular momentum, we define
the vector angular momentum L about a point as the vector moment of
the momentum vector about 0:
L = r X p = m(r x v), (3-142)
where the vector r is taken from the point as origin to the position of the
particle of mass m. Again we shall omit the subscript when no confusion
can arise. The component of the vector L in any direction is the moment
of the momentum vector p about an axis in that direction through 0.
By taking the cross product of r with both members of the vector equa-
tion of motion [Eq. (3-125)], we obtain
By the rules of vector algebra and vector calculus,
dL d r , ,,
It = dt [T * (mv)]
= r X j t (mv) + v x (mv)
-■"•(-S)-
We substitute this result in Eq. (3-143) :
dL/dt = r X F == N. (3-144)
The time rate of change of the vector angular momentum of a particle is
equal to the vector torque acting on it. The integral form of the angular
momentum theorem is
(3-145)
The theorems for plane angular momentum and for angular momentum
about an axis follow from the vector angular momentum theorems by tak-
ing components in the appropriate direction.
104 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
3-9 Discussion of the general problem of two- and three-dimensional
motion. If the force F is given, in general as a function F(v, r, t) of position,
velocity, and time, the equations of motion (3-126) become a set of three
(or, in two dimensions, two) simultaneous second-order differential equa-
tions:
d 2 x
m -Tp = F x (x, y, z, x, y, z, f),
m § = F v (x, y,z,x,y,z,t), (3-146)
dP
dh
= F z (x, y, z, x, y, z, t).
If we are given the position r = (x , 2/o, Zo), and the the velocity v =
(v X0 , v yo , v Z0 ) at any instant t , Eqs. (3-146) give us d 2 r/dt 2 , and from r, f , f ,
at time t, we can determine r, f a short time later or earlier at t + dt, thus
extending the functions r, f , f, into the past and future with the help of
Eqs. (3-146). This argument can be made mathematically rigorous, and
leads to an existence theorem guaranteeing the existence of a unique solu-
tion of these equations for any given position and velocity at an initial
instant t . We note that the general solution of Eqs. (3-146) involves the
six "arbitrary" constants x , yo, z , v xo , v yo , v 20 . Instead of these six con-
stants, we might specify any other six quantities from which they can be
determined. (In two dimensions, we will have two second-order differential
equations and four initial constants.)
In general, the solution of the three simultaneous equations (3-146)
will be much more difficult than the solution of the single equation (2-9)
for one-dimensional motion. The reason for the greater difficulty is that,
in general, all the variables x, y, z and their derivatives are involved in all
three equations, which makes the problem of the same order of difficulty
as a single sixth-order differential equation. [In fact, the set of Eqs. (3-146)
can be shown to be equivalent to a single sixth-order equation.] If each
force component involved only the corresponding coordinate and its deriva-
tives,
F x = F x (x, x, t),
F v = F v (y, y, t), (3-147)
F z = F z {z, z, t),
then the three equations (3-146) would be independent of one another.
We could solve for x(t), y{t), z{t) separately as three independent problems
in one-dimensional motion. The most important example of this case is
3-9] DISCUSSION OF THE GENERAL PROBLEM 105
probably when the force is given as a function of time only:
F = F«) = [F x (t), F v (t), FM- (3-148)
The x, y, and z equations of motion car} then each be solved separately by
the method given in Section 2-3. The case of a frictional force propor-
tional to the velocity will also be an example of the type (3-147). Other
cases will sometimes occur, for example, the three-dimensional harmonic
oscillator (e.g., a baseball in a tubful of gelatine, or an atom in a crystal
lattice), for which the force is
F y = —k y y, (3-149)
F z = —k z z,
when the axes are suitably chosen. The problem now splits into three
separate linear harmonic oscillator problems in x, y, and z. In most cases,
however, we are not so fortunate, and Eq. (3-147) does not hold. Special
methods are available for solving certain classes of two- and three-dimen-
sional problems. Some of these will be developed in this chapter. Prob-
lems not solvable by such methods are always, in principle, solvable by
various numerical methods of integrating sets of equations like Eqs. (3-146)
to get approximate solutions to any required degree of accuracy. Such
methods are even more tedious in the three-dimensional case than in the
one-dimensional case, and are usually impractical unless one has the
services of one of the large automatic computing machines.
When we try to extend the idea of potential energy to two or three
dimensions, we will find that having the force given as F(r), a function of
r alone, is not sufficient to guarantee the existence of a potential-energy
function V(t). In the one-dimensional case, we found that if the force is
given as a function of position alone, a potential-energy function can al-
ways be defined by Eq. (2-41). Essentially, the reason is that in one
dimension, a particle which travels from xi to x 2 and returns to X\ must
return by the same route, so that if the force is a function of position alone,
the work done by the force on the particle during its return trip must nec-
essarily be the same as that expended against the force in going from xi
to x 2 . In three dimensions, a particle can travel from rj to r 2 and return
by a different route, so that even if F is a function of r, the particle may be
acted on by a different force on the return trip and the work done may
not be the same. In Section 3-12 we shall formulate a criterion to determine
when a potential energy V(t) exists.
When V(r) exists, a conservation of energy theorem still holds, and the
total energy (T + V) is a constant of the motion. However, whereas in
106 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
one dimension the energy integral is always sufficient to enable us to solve
the problem at least in principle (Section 2-5), in two and three dimensions
this is no longer the case. If x is the only coordinate, then if we know a
relation (T + V = E) between x and x, we can solve for x = f(x) and
reduce the problem to one of carrying out a single integration. But with
coordinates x, y, z, one relation between x, y, z, x, y, z is not enough. We
would need to know five such relations, in general, in order to eliminate, for
example, x, y, x, and y, and find z = f(z). In the two-dimensional case, we
would need three relations between x, y, x, y to solve the problem by this
method. To find four more relations like the energy integral from Eqs.
(3-146) (or two more in two dimensions) is hopeless in most cases. In
fact, such relations do not usually exist. Often, however, we can find other
quantities (e.g., the angular momentum) which are constants of the motion,
and thus obtain one or two more relations between x, y, z, x, y, z, which in
many cases will be enough to allow a solution of the problem. Examples
will be given later.
3-10 The harmonic oscillator in two and three dimensions. In this
section and the next, we consider a few simple problems in which the force
has the form of Eqs. (3-147), so that the equations of motion separate into
independent equations in x, y, and z. Mathematically, we then simply
have three separate problems, each of the type considered in Chapter 2.
The only new feature will be the interpretation of the three solutions
x(t), y(t), z(t) as representing a motion in three-dimensional space.
We first consider briefly the solution of the problem of the three-dimen-
sional harmonic oscillator without damping, whose equations of motion
are
my = -k v y, (3-150)
mz = —k z z.
A model could be constructed by suspending a mass between three per-
pendicular sets of springs (Fig. 3-27). The solutions of these equations,
we know from Section 2-8:
x = A x cos (a x t + 6 X ), o x = k x /m,
y = A y cos (o)yt + By), w| = ky/m, (3-151)
z = A z cos (u z t + e z ), oil = kz/m.
The six constants (A x , A y , A z , d x , 6 y , d z ) depend on the initial values
£o, 2/o, zo, %o, Vo, 2o- Each coordinate oscillates independently with simple
3-10]
THE HARMONIC OSCILLATOR
107
Fig. 3-27. Model of a three-dimensional harmonic oscillator.
harmonic motion at a frequency depending on the corresponding restoring
force coefficient, and on the mass. The resulting motion of the particle
takes place within a rectangular box of dimensions 2A X X 2A y X 2A t
about the origin. If the angular frequencies w xi u v , w z are commensurable,
that is, if for some set of integers (n x , n^, n x ),
n x
U>y
Ml
(3-152)
then the path of the mass m in space is closed, and the motion is periodic.
If (n m n y , n a ) are chosen so that they have no common integral factor,
then the period of the motion is
t =
27171,
2irn y
2-irrit
W,T
(3-153)
During one period, the coordinate x makes n x oscillations, the coordinate y
makes % oscillations, and the coordinate z makes n* oscillations, so that
the particle returns at the end of the period to its initial position and
velocity. In the two-dimensional case, if the path of the oscillating parti-
cle is plotted for various combinations of frequencies^ and «„, and various
phases B x and 9 V , many interesting and beautiful patterns are obtained.
Such patterns are called Lissajous figures (Fig. 3-28), and may be pro-
duced mechanically by a mechanism designed to move a pencil or other
writing device according to Eqs. (3-151). Similar patterns may be ob-
tained electrically on a cathode-ray oscilloscope by sweeping horizontally
108 MOTION OF PABTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3
Fig. 3-28. Lissajous figures.
and vertically with suitable oscillating voltages. If the frequencies u x ,
u y , u z are incommensurable, so that Eq. (3-152) does not hold for any set
of integers, the motion is not periodic, and the path fills the entire box
2A X X 2A V X 2A Z , in the sense that the particle eventually comes ar-
bitrarily close to every point in the box. The discussion can readily be
extended to the cases of damped and forced oscillations in two and three
dimensions.
If the three constants k x , k v , k z are all equal, the oscillator is said to be
isotropic, that is, the same in all directions. In this case, the three fre-
quencies u x , u y , a z are all equal and the motion is periodic, with each coordi-
nate executing one cycle of oscillation in a period. The path can be shown
to be an ellipse, a straight line, or a circle, depending on the amplitudes
and phases (A x , A y , A z , d x , 6 y , B z ).
3-11 Projectiles. An important problem in the history of the science of
mechanics is that of determining the motion of a projectile. A projectile
moving under the action of gravity near the surface of the earth moves, if
3 11] PROJECTILES 109
air resistance is neglected, according to the equation
d 2 x
m dj2 = -™#K (3-154)
where the 2-axis is taken in the vertical direction. In component form:
d 2 x
™^f = 0, (3-155)
»^=0, (3-156)
d 2 z
m ^2 = —m- (3-157)
The solutions of these equations are
x — x + v xo t, (3-158)
V = 2/o + » w «, (3-159)
z = 2 + *>*„* — k< 2 , (3-160)
or, in vector form,
r = r + v t - igt 2 k. (3-161)
We assume the projectile starts from the origin (0, 0, 0), with its initial
velocity in the as-plane, so that v yo = 0. This is no limitation on the
motion of the projectile, but merely corresponds to a convenient choice of
coordinate system. Equations (3-158), (3-159), (3-160) then become
* = Vx t, (3-162)
V = 0, (3-163)
z = Vzo t — \gt 2 . (3-164)
These equations give a complete description of the motion of the projectile.
Solving the first equation for t and substituting in the third, we have an
equation for the path in the zz-plane:
2 = — * -i-f-z 2 . (3-165)
This can be rewritten in the form
110 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
This is a parabola, concave downward, whose maximum altitude occurs at
2
v
2g
z m = 7~' (3-167)
and which crosses the horizontal plane z = at the origin and at the point
2 JW^o. (3 _ 168)
9
If the surface of the earth is horizontal, x m is the range of the projectile.
Let us now take account of air resistance by assuming a frictional force
proportional to the velocity:
m g = _ m?k _ 6 |. ( 3 - 169 )
In component notation, if we assume that the motion takes place in the
a;2-plane,
m§=-hf t > 0-170)
m w* = ~ m - 6 i' (3_171)
It should be pointed out that the actual resistance of the air against a mov-
ing projectile is a complicated function of velocity, so that the solutions we
obtain will be only approximate, although they indicate the general nature
of the motion. If the projectile starts from the origin at t = 0, the solu-
tions of Eqs. (3-170) and (3-171) are (see Sections 2-4 and 2-6)
,. _ ,. --Kim
(3-172)
x = m^o (1 _ e -M/») f
(3-173)
^ = (x + ^) e " 6i/m -?'
(3-174)
• - Op* + nr) c 1 - e ~ ulm)
■mg ,
b *•
(3-175)
Solving Eq. (3-173) for t and substituting in Eq. (3-175), we obtain an
equation for the trajectory:
■ -(£ + ?*)' -# ta (=r^ta)- «» 76 >
3-11]
PROJECTILES HI
^^rr
Fig. 3-29. Trajectories for maximum range for projectiles with various muzzle
velocities.
For low air resistance, or short distances, when (bx)/(mv Xtt ) « 1, we may
expand in powers of (bx)/(mv xo ) to obtain
*-^*-*i* 2 -*-^* 3 ----. (3 _ 177)
v x v XQ mvt
Thus the trajectory starts out as a parabola, but for larger values of x
(taking v Xo as positive), z falls more rapidly than for a parabola. Accord-
ing to Eq. (3-176), as x approaches the value (mv xo )/b, z approaches minus
infinity, i.e., the trajectory ends as a vertical drop at x = (mv xo )/b. From
Eq. (3-174), we see that the vertical fall at the end of the trajectory takes
place at the terminal velocity —mg/b. (The projectile may, of course,
return to earth before reaching this part of its trajectory.) If we take the
first three terms in Eq. (3-177) and solve for x when z = 0, we have ap-
proximately, if aw « (mv X0 )/b,
fc ***»-l*£ + .... (3-m)
The second term gives the first-order correction to the range due to air re-
sistance, and the first two terms will give a good approximation when the
effect of air resistance is small. The extreme opposite case, when air
resistance is predominant in determining range (Fig. 3-29), occurs when
the vertical drop at x = (mv xo )/b begins above the horizontal plane z = 0.
The range is then, approximately,
b \mg /
mg / < 3 - 179 )
We can treat (approximately) the problem of the effect of wind on the
projectile by assuming the force of air resistance to be proportional to
the relative velocity of the projectile with respect to the air:
m d£= -Wl k - b (§ - v ») ' (3-180)
where v w is the wind velocity. If v„ is constant, the term 6v„, in Eq. (3-180)
112 MOTION OF PARTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3
behaves as a constant force added to -mgk, and the problem is easily
solved by the method above, the only difference being that there may be
constant forces in addition to frictional forces in all three directions x, y, z.
The air resistance to a projectile decreases with altitude, so that a better
form for the equation of motion of a projectile which rises to appreciable
altitudes would be
where h is the height (say about five miles) at which the air resistance falls
to 1/e of its value at the surface of the earth. In component form,
mx = —b±e~" h , my = -bye-" h , (3-182)
ml = —mg — bze~" .
These equations are much harder to solve. Since z appears in the x and
y equations, we must first solve the z equation for z(t) and substitute in
the other two equations. The z equation is not of any of the simple types
discussed in Chapter 2. The importance of this problem was brought out
during the First World War, when it was discovered accidentally that
aiming a cannon at a much higher elevation than that which had previ-
ously been believed to give maximum range resulted in a great increase in
the range of the shell. The reason is that the reduction in air resistance,
at altitudes of several miles, more than makes up for the loss in horizontal
component of muzzle velocity resulting from aiming the gun higher.
3-12 Potential energy. If the force F acting on a particle is a function
of its position r = (x, y, z), then the work done by the force when the
particle moves from ri to r 2 is given by the line integral
f
Jii
F-dr.
It is suggested that we try to define a potential energy 7(r) = V(x, y, z) in
analogy with Eq. (2^11) for one-dimensional motion, as the work done by
the force on the particle when it moves from r to some chosen standard
point r g :
V(i) = - f F(r)-dr. (3-183)
Such a definition implies, however, that the function V(i) shall be a func-
tion only of the coordinates (x, y, z) of the point r (and of the standard
point r s , which we regard as fixed), whereas in general the integral on the
3 _ 12] POTENTIAL ENERGY
113
right depends upon the path of integration from r, to r. Only if the integral
on the right is independent of the path of integration will the definition
be legitimate.
Let us assume that we have a force function F(x, y, z) such that the line
integral in Eq. (3-183) is independent of the path of integration from r s to
any point r. The value of the integral then depends only on r (and on r s ),
and Eq. (3-183) defines a potential energy function V(r). The change in
V when the particle moves from r to r + dr is the negative of the work
done by the force F:
dV = —F'dr. (3-184)
Comparing Eq. (3-184) with the geometrical definition [Eq. (3-107)] of
the gradient, we see that
— F = grad V,
F « -W. (3 " 185)
Equation (3-185) may be regarded as the solution of Eq. (3-183) for F in
terms of V. In component form,
'--£■ '--%■ '--£■ <MW
In seeking a condition to be satisfied by the function F(r) in order that
the integral in Eq. (3-183) be independent of the path, we note that, since
Eq. (3-28) can be proved from the algebraic definition of the cross product,
it must hold also for the vector symbol V:
V X V = 0. (3-187)
Applying (V X V) to the function V, we have
V X W = curl (grad 7) = 0. (3-188)
Equation (3-188) can readily be verified by direct computation. From
Eqs. (3-188) and (3-185), we have
V X F = curl F = 0. (3-189)
Since Eq. (3-189) has been deduced on the assumption that a potential
function exists, it represents a necessary condition which must be satisfied
by the force function F(x, y, z) before a potential function can be defined.
We can show that Eq. (3-189) is also a sufficient condition for the existence
of a potential by making use of Stokes' theorem [Eq. (3-117)]. By Stokes'
theorem, if we consider any closed path C in space, the work done by the
114 MOTION OF PARTICLE IN TWO OE THREE DIMENSIONS [CHAP. 3
*2
ri
Fig. 3-30. Two paths between n and r 2 , forming a closed path.
force F(r) when the particle travels around this path is
f F-di = /7n-(V x F) dS, (3-190)
s
where S is a surface in space bounded by the closed curve C. If now
Eq. (3-189) is assumed to hold, the integral on the right is zero, and we
have, for any closed path. C,
/«
F-dt = 0. (3-191)
c
But if the work done by the force F around any closed path is zero, then
the work done in going from r x to r 2 will be independent of the path fol-
lowed. For consider any two paths between r x and r 2 , and let a closed
path C be formed going from r x to r 2 by one path and returning to ri
by the other (Fig. 3-30). Since the work done around C is zero, the work
going from r x to r 2 must be equal and opposite to that on the return trip,
hence the work in going from ri to r 2 by either path is the same. Applying
this argument to the integral on the right in Eq. (3-183), we see that the
result is independent of the path of integration from r s to r, and therefore
the integral is a function V(r) of the upper limit alone, when the lower
limit r, is fixed. Thus Eq. (3-189) is both necessary and sufficient for the
existence of a potential function V(r) when the force is given as a function
of position F(r).
When curl F is zero, we can express the work done by the force when
the particle moves from r x to r 2 as the difference between the values of the
potential energy at these points:
pF.dr= r-F-dt+TF-dr
Jti Jti Ji,
= V(ri) - F(r 9 ). (3-192)
Combining Eq. (3-192) with the energy theorem (3-135), we have for any
two times ti and Z 2 :
T x + F(r0 = T 2 + V(r 2 ). (3-193)
3-12] POTENTIAL ENERGY 115
Hence the total energy (T + V) is again constant, and we have an energy
integral for motion in three dimensions:
T + V = %m(x 2 + y 2 + z 2 ) + V(x, y, z) = E. (3-194)
A force which is a function of position alone, and whose curl vanishes, is
said to be conservative, because it leads to the theorem of conservation of
kinetic plus potential energy [Eq. (3-194)].
In some cases, a force may be a function of both position and time
F(r, t). If at any time t the curl of F(r, t) vanishes, then a potential-energy
function V(r, t) can be defined as
V(i,t) = - f F(r,«)-dr, (3-195)
and we will have, for any time t such that V X F(r, t) = 0,
F(r, t) = -VF(r, t). (3-196)
However, the conservation law of energy can no longer be proved, for
Eq. (3-192) no longer holds. It is no longer true that the change in po-
tential energy equals the negative of the work done on the particle, for
the integral which defines the potential energy at time * is computed from
the force function at that time, whereas the integral that defines the work
is computed using at each point the force function at the time the particle
passed through that point. Consequently, the energy T + V is not a
constant when F and V are functions of time, and such a force is not to
be called a conservative force.
When the forces acting on a particle are conservative, Eq. (3-194)
enables us to compute its speed as a function of its position. The energy E
is fixed by the initial conditions of the motion. Equation (3-194), like
Eq. (2-44), gives no information as to the direction of motion. This lack
of knowledge of direction is much more serious in two and three dimensions,
where there is an infinity of possible directions, than in one dimension,
where there are only two opposite directions in which the particle may
move. In one dimension, there is only one path along which the particle
may move. In two or three dimensions, there are many paths, and unless
we know the path of the particle, Eq. (3-194) alone allows us to say very
little about the motion except that it can occur only in the region where
V{x, y, z) < E. As an example, the potential energy of an electron in the
attractive electric field of two protons (ionized hydrogen molecule H 2 + ) is
2 2
v = ~ k ~ k ' (esu) (3_197)
116 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
-18
Fig. 3-31. Potential energy of electron in electric field of two protons 2 A
apart. (Potential energies in units of 10 -12 erg.)
where r\, r 2 are the distances of the electron from the two protons. The
function V(x, y) (for motion in the x2/-plane only) is plotted in Fig. 3-31
as a contour map, where the two protons are 2 A apart at the points y = 0,
x = ±1 A, and the figures on the contours of constant potential energy are
the corresponding potential energies in units of 10 -12 erg. So long as
E < —46 X 10~ 12 erg, the electron is confined to a region around one
proton or the other, and we expect its motion will be either an oscillation
through the attracting center or an orbit around it, depending on initial
conditions. (These comments on the expected motion require some physical
insight or experience in addition to what we can say from the energy integral
alone.) For > E > —46 X 10 -12 erg, the electron is confined to a
region which includes both protons, and a variety of motions are possible.
For E > 0, the electron is not confined to any finite region in the plane.
For E <<C —46 X 10~ 12 erg, the electron is confined to a region where the
equipotentials are nearly circles about one proton, and its motion will be
practically the same as if the other proton were not there. For E < 0,
but \E\ <5C 46 X 10 -12 erg, the electron may circle in an orbit far from the
attracting centers, and its motion then will be approximately that of an
electron bound to a single attracting center of charge 2e, as the equi-
potential lines far from the attracting centers are again very nearly circles.
3-12]
POTENTIAL ENERGY
117
Given a potential energy function V(x, y, z), Eq. (3-186) enables us to
compute the components of the corresponding force at any point. Con-
versely, given a force F(x, y, z), we may compute its curl to determine
whether a potential energy function exists for it. If all components of
curl F are zero within any region of space, then within that region, F may
be represented in terms of a potential-energy function as —VV. The
potential energy is to be computed from Eq. (3-183). Furthermore, since
curl F = 0, the result is independent of the path of integration, and we
may compute the integral along any convenient path. As an example,
consider the following two force functions:
(a) F x = axy,
Fy = -az 2 ,
F,= -
ax
(b) F x = ay(y 2 - 3* 2 ), F y = 3ax(y 2 - z 2 ), F z = ~6axyz,
where a is a constant. We compute the curl in each case:
(a)
1 \dy dz/^ J \dz dx/ + K \dx dyj
= (2az)i + (2ax)j - (ax)k,
(b) V X F = 0.
In case (a) no potential energy exists. In case (b) there is a potential
energy function, and we proceed to find it. Let us take r„ = 0, i.e., take
the potential as zero at the origin. Since the components of force are given
as functions of x, y, z, the simplest path of integration from (0, 0, 0) to
(#0, Vo, zo) along which to compute the integral in Eq. (3-183) is one which
follows lines parallel to the coordinate axes, for example as shown in
Fig. 3-32:
V(x ,y ,z ) = — f xo,Vo ' z ° p-dr = — f F-di — f F-dr — f F-dr.
J(0,0,0) JCi Jc 2 Jc 3
(
•(xo,yo,zo)
,
c 3
(0,0,0)
y
(nU)/
c 2
Oo,W>,0)
Fig. 3-32. A path of integration from (0, 0, 0) to (xo, Vo, «o).
118 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
Now along Ci, we have
y = z = 0, F x = F y = F z = 0, dr = i da;.
Thus
/" F-dr = f x °F x dx = 0.
JCi JO
Along C 2 ,
a; — X0)
z =
o,
Thus
F*
= ay
3 , Fy =
dr =
Zax Q y'
\dy.
, F,=
= 0,
J. 11U.O
Along
c 3)
Jc 2
F-dr = /
Jo
X = Xq,
Fydy
y =
= ax y%-
yo,
F x
= 02/0(2/0
—
3* 2 ),
Fy — Sax (yl
-z\
F z
Thus
dr =
kdz.
_2
-6axo2/oz,
f F«dr = f " F z dz = —Zax yoZ 2 .
JCs Jo
Thus the potential energy, if the subscript zero is dropped, is
V(x, y, z) = —axy 3 + Saxyz 2 .
It is readily verified that the gradient of this function is the force given by
(b) above. In fact, one way to find the potential energy, which is often
faster than the above procedure, is simply to try to guess a function whose
gradient will give the required force.
An important case of a conservative force is the central force, a force
directed always toward or away from a fixed center O, and whose magni-
tude is a function only of the distance from 0. In spherical coordinates,
with as origin,
F = nF(r). (3-198)
The cartesian components of a central force are (since n = r/r)
F x = ^F(r),
F v = y - F(x), [r = (x 2 + y 2 + z 2 ) 112 ], (3-199)
T
F z = *-F(r).
3-12]
POTENTIAL ENERGY
119
(ro,8o,<po)
(r,,6 s ,ip s )
Fig. 3-33. Path of integration for a central force.
The curl of this force can be shown by direct computation to be zero, no
matter what the function F(r) may be. For example, we find
*** = x d ( F ( r A dr _ X V d (F(r)\
dy dr\ r ) by r dr\ r ) '
= d (F(r)\ dr ^ xy d (F(r)\
dr\ r ) dx r dr\ r )
111
dx
Therefore the g-component of curl F vanishes, and so, likewise, do the other
two components. To compute the potential energy, we choose any stand-
ard point r s , and integrate from r 8 to r along a path (Fig. 3-33) following
a radius (Ci) from i„ whose coordinates are (r„ 0„ <p,), to the point
0"o, 0«, <P»), then along a circle (C 2 ) of radius r about the origin to the
point (r , e 0) <po). Along C u
dr = n dr,
Along C 2 ,
Thus
f F-dr = f ° F(r) dr.
JCi Jr,
dt = ]r dd + tor sin d<p,
f F-dr = 0.
Jc 2
V(to) = - r ?-dr = - f F-dr - f F-
Jr, JCi JC 2
= - f° F(r) dr.
The potential energy is a function of r alone:
V(t) = V(r) = - f F(r) dr.
Jr,
dr
(3-200)
120 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
3-13 Motion under a central force. A central force is a force of the form
given by Eq. (3-198). Physically, such a force represents an attraction
[if F(r) < 0] or repulsion [if F(r) > 0] from a fixed point located at the
origin r = 0. In most cases where two particles interact with each other,
the force between them is (at least primarily) a central force; that is, if
either particle be located at the origin, the force on the other is given by
Eq. (3-198). Examples of attractive central forces are the gravitational
force acting on a planet due to the sun, or the electrical attraction acting
on an electron due to the nucleus of an atom. The force between a proton
or an alpha particle and another nucleus is a repulsive central force. In
the most important cases, the force F(r) is inversely proportional to r 2 .
This case will be treated in the next section. Other forms of the function
F(r) occur occasionally; for example, in some problems involving the struc-
ture and interactions of nuclei, complex atoms, and molecules. In this
section, we present the general method of attack on the problem of a par-
ticle moving under the action of a central force.
Since in all these examples, neither of the two interacting particles
is actually fastened to a fixed position, the problem we are solving, like
most problems in physics, represents an idealization of the actual problem,
valid when one of the particles can be regarded as practically at rest at
the origin. This will be the case if one of the particles is much heavier
than the other. Since the forces acting on the two particles have the same
magnitude by Newton's third law, the acceleration of the heavy one will
be much smaller than that of the lighter one, and the motion of the heavy
particle can be neglected in comparison with the motion of the lighter one.
We shall discover later, in Section 4-7, that, with a slight modification, our
solution can be made to yield an exact solution to the problem of the
motion of two interacting particles, even when their masses are equal.
We may note that the vector angular momentum of a particle under
the action of a central force is constant, since the torque is
N = rxF=(rx n)F(r) = 0. (3-201)
Therefore, by Eq. (3-144),
f = 0. (3-202)
As a consequence, the angular momentum about any axis through the
center of force is constant. It is because many physical forces are central
forces that the concept of angular momentum is of importance.
In solving for the motion of a particle acted on by a central force, we
first show that the path of the particle lies in a single plane containing the
center of force. To show this, let the position r and velocity v<> be given
at any initial time t , and choose the rc-axis through the initial position r
3-13] MOTION UNDER A CENTRAL FORCE 121
of the particle, and the z-axis perpendicular to the initial velocity v . Then
we have initially:
*o = jr |, 2/o = zo = 0, (3-203)
Vx = v -i, v yo = v -j, v ls> = 0. (3-204)
The equations of motion in rectangular coordinates are, by Eqs. (3-199),
mx = j F(r), my = ^ F(r), mz = - F(r). (3-205)
A solution of the z-equation which satisfies the initial conditions on z and
Vz is
z(t) = 0. (3-206)
Hence the motion takes place entirely in the a;j/-plane. We can see phys-
ically that if the force on a particle is always toward the origin, the particle
can never acquire any component of velocity out of the plane in which it
is initially moving. We can also regard this result as a consequence of
the conservation of angular momentum. By Eq. (3-202), the vector
L — m(i X v) is constant; therefore both r and v must always lie in a
fixed plane perpendicular to L.
We have now reduced the problem to one of motion in a plane with two
differential equations and four initial conditions remaining to be satisfied.
If we choose polar coordinates r, in the plane of the motion, the equations
of motion in the r and directions are, by Eqs. (3-80) and (3-198),
mr — mrd 2 = F(r), (3-207)
mr8 + 2mr& = 0. (3-208)
Multiplying Eq. (3-208) by r, as in the derivation of the (plane) angular
momentum theorem, we have
it ^» = d -tt = °- ( 3 - 209 >
This equation expresses the conservation of angular momentum about the
origin and is a consequence also of Eq. (3-202) above. It may be inte-
grated to give the angular momentum integral of the equations of motion:
mr 2 d = L = a constant. (3-210)
The constant L is to be evaluated from the initial conditions. Another
integral of Eqs. (3-207) and (3-208), since the force is conservative, is
T + V = Imr* + imr 2 6 2 + V(r) = E, (3-211)
122 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
where V(r) is given by Eq. (3-200) and E is the energy constant, to be
evaluated from the initial conditions. If we substitute for 6 from Eq.
(3-210), the energy becomes
hmf 2 + £2 + V(r) = E. (3-212)
We can solve for f:
^I^-^-SfT- (3 - 2i3)
Therefore
/
dr = Jft- t. (3-214)
/ / t2 \i/2 -\ m
[ [ E - ™ - £^)
The integral is to be evaluated and the resulting equation solved for r(t).
We then obtain 0(0 from Eq. (3-210):
' = '« + /„' Ji dt - < 3 - 215 >
We thus obtain the solution of Eqs. (3-207) and (3-208) in terms of the
four constants L, E, r , do, which can be evaluated when the initial position
and velocity in the plane are given.
It will be noted that our treatment based on Eq. (3-212) is analogous
to our treatment of the one-dimensional problem based on the energy in-
tegral [Eq. (2-44)]. The coordinate r here plays the role of x, and the 6
term in the kinetic energy, when 6 is eliminated by Eq. (3-210), plays the
role of an addition to the potential energy. We may bring out this analogy
further by substituting from Eq. (3-210) into Eq. (3-207):
mf - -^ = F(r). (3-216)
If we transpose the term —L 2 /mr 3 to the right side, we obtain
mr = F(r)+£- 3 - (3-217)
This equation has exactly the form of an equation of motion in one dimen-
sion for a particle subject to the actual force F(r) plus a "centrifugal force"
L 2 /mr 3 . The centrifugal force is not really a force at all but a part of the
mass times acceleration, transposed to the right side of the equation in
order to reduce the equation for r to an equation of the same form as for
one-dimensional motion. We may call it a "fictitious force." If we treat
3-13] MOTION UNDER A CENTRAL FORCE 123
Eq. (3-217) as a problem in one-dimensional motion, the effective "poten-
tial energy" corresponding to the "force" on the right is
>V>(r) = -f m dr-f£.dr
The second term in 'V is the "potential energy" associated with the
"centrifugal force." The resulting energy integral is just Eq. (3-212).
The reason why we have been able to obtain a complete solution to our
problem based on only two integrals, or constants of the motion (L and E),
is that the equations of motion do not contain the coordinate 0, so that the
constancy of L is sufficient to enable us to eliminate entirely from Eq.
(3-207) and to reduce the problem to an equivalent problem in one-dimen-
sional motion.
The integral in Eq. (3-214) sometimes turns out rather difficult to eval-
uate in practice, and the resulting equation difficult to solve for r(t). It is
sometimes easier to find the path of the particle in space than to find its
motion as a function of time. We can describe the path of the particle by
giving r(0). The resulting equation is somewhat simpler if we make the
substitution
(3-219)
1
u = - 1
r
1
r = -•
u
Then
we
have,
using
r
Eq. (3-210),
_ 1 du ,
«2 dd
_ L du
m de
= -r 2 6^
de
f
L d 2 u *
~ m~cW
r2„,2
L u
d 2 u
de 2
(3-220)
(3-221)
Substituting for r and f in Eq. (3-217), and multiplying by —m/(L 2 u 2 ), we
have a differential equation for the path or orbit in terms of u(fi) :
In case L = 0, Eq. (3-222) blows up, but we see from Eq. (3-210) that in
this case 6 is constant, and the path is a straight line through the origin.
124 MOTION OF PARTICLE IN TWO OK THBEE DIMENSIONS [CHAP. 3
Even in cases where the explicit solutions of Eqs. (3-214) and (3-215),
or Eq. (3-222), are difficult to carry through, we can obtain qualitative
information about the r motion from the effective potential 'V given by
Eq. (3-218), just as in the one-dimensional case discussed in Section 2-5.
By plotting 'V'(r), we can decide for any total energy E whether the
motion in r is periodic or aperiodic, we can locate the turning points, and
we can describe roughly how the velocity f varies during the motion. If
'V'(r) has a minimum at a point r , then for energy E slightly greater than
'V (r ), r may execute small, approximately harmonic oscillations about r
with angular frequency given by
CO
_ 1 (d 2 'V'\
m \ dr 2 /
2 = ^ ^ • (3-223)
[See the discussion in Section 2-7 concerning Eq. (2-87).] We must re-
member, of course, that at the same time the particle is revolving around
the center of force with an angular velocity
i = ~ (3-224)
mr 2
The rate of revolution decreases as r increases. When the r motion is
periodic, the period of the r motion is not, in general, the same as the
period of revolution, so that the orbit may not be closed, although it is
confined to a finite region of space. (See Fig. 3-34.) In cases where the
r motion is not periodic, then 6 — > as r — * oo , and the particle may or
may not perform one or more complete revolutions as it moves toward
r = oo , depending on how rapidly r increases. In the event the motion is
periodic, that is, when the particle moves in a closed orbit, the period of
orbital motion is related to the area of the orbit. This can be seen as
follows. The area swept out by the radius from the origin to the particle
Fig. 3-34. An aperiodic bounded orbit.
3~14] INVERSE SQUARE LAW FORCE 125
Fig. 3-35. Area swept out by radius vector.
when the particle moves through a small angle dd is approximately
(Fig. 3-35)
dS = %r 2 dd. (3-225)
Hence the rate at which area is swept out by the radius is, by Eq. (3-210),
^12/ L
Tt = ^ 6 = 2^- ( 3 - 22 6)
This result is true for any particle moving under the action of a central
force. If the motion is periodic, then, integrating over a complete period r
of the motion, we have for the area of the orbit
S = 2^ * < 3 - 227 )
If the orbit is known, the period of revolution can be calculated from this
formula.
3-14 The central force inversely proportional to the square of the
distance. The most important problem in three-dimensional motion is
that of a mass moving under the action of a central force inversely pro-
portional to the square of the distance from the center:
F = f 2 n, (3-228)
for which the potential energy is
V(r) = j , (3-229)
where the standard radius r s is taken to be infinite in order to avoid an
additional constant term in V(r). As an example, the gravitational force
(Section 1-5) between two masses mi and m 2 a distance r apart is given by
Eq. (3-228) with
K = —Gm y m 2 , G = 6.67 X 10 -8 dyne-gm -2 -cm 2 , (3-230)
where K is negative, since the gravitational force is attractive. Another
126 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
•V\r)
-\{Khn/V)
Fig. 3-36. Effective potential for central inverse square law of force.
example is the electrostatic force between two electric charges gi and q 2 a
distance r apart, given by Eq. (3-228) with
K = qiq 2 ,
(3-231)
where the charges are in electrostatic units, and the force is in dynes. The
electrostatic force is repulsive when qi and q 2 have the same sign, other-
wise attractive. Historically, the first problems to which Newton's me-
chanics was applied were problems involving the motion of the planets
under the gravitational attraction of the sun, and the motion of satellites
around the planets. The success of the theory in accounting for such
motions was responsible for its initial acceptance.
We first determine the nature of the orbits given by the inverse square
law of force. In Fig. 3-36 is plotted the effective potential
'V'(r) = — + -=—■
(3-232)
For a repulsive force (K > 0), there are no periodic motions in r; only
positive total energies E are possible, and the particle comes in from r = oo
to a turning point and travels out to infinity again. For a given energy
and angular momentum, the turning point occurs at a larger value of r than
for K = (no force"), for which the orbit would be a straight line. For an
attractive force (K < 0) with L^O, the motion is also unbounded if
E > 0, but in this case the turning point occurs at a smaller value of r than
for K = 0. Hence the orbits are as indicated in Fig. 3-37. The light
3-14]
INVEBSE SQUABE LAW FOBCE
127
Fig. 3-37. Sketch of unbounded inverse square law orbits.
lines in Fig. 3-37 represent the turning point radius or perihelion distance
measured from the point of closest approach of the particle to the attracting
or repelling center. For K < 0, and -\K 2 m/L 2 < E < 0, the coordinate
r oscillates between two turning points. For.E= —\K 2 m/h 2 , the particle
moves in a circle of radius r = L 2 /(—Km). Computation shows (see
Problem 30 at the end of this chapter) that the period of small oscillations
in r is the same as the period of revolution, so that for E near — \K 2 m/L 2 ,
the orbit is a closed curve with the origin slightly off center. We shall show
later that the orbit is, in fact, an ellipse for all negative values of E if
L ?* 0. If L = 0, the problem reduces to the one-dimensional motion of
a falling body, discussed in Section 2-6.
To evaluate the integrals in Eqs. (3-214) and (3-215) for the inverse
square law of force is rather laborious. We shall find that we can obtain all
the essential information about the motion more simply by starting from
Eq. (3-222) for the orbit. Equation (3-222) for the orbit becomes, in this
case,
d 2 u . mK
-j^ + u= - — ■ (3-233)
d0 2
L2
This equation has the same form as that of a harmonic oscillator (of unit
frequency) subject to a constant force, where B here plays the role of t.
The homogeneous equation and its general solution are
d 2 u .
(3-234)
u = A cos (0 — d ),
(3-235)
128 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
where A, O are arbitrary constants. An obvious particular solimon of the
inhomogeneous equation (3-233) is the constant solution
„=_*££ (3-236)
Hence the general solution of Eq. (3-233) is
u = - = - ^ + A cos (0 - O ). (3-237)
r L 2
This is the equation of a conic section (ellipse, parabola, or hyperbola) with
focus at r = 0, as we shall presently show. The constant O determines
the. orientation of the orbit in the plane. The constant A, which may be
taken as positive (since O is arbitrary), determines the turning points of
the r motion, which are given by
1 = _^ + A) ! = _^_ A . (3-238)
If A > —mK/L 2 (as it necessarily is for K > 0), then there is only one
turning point, ri, since r cannot be negative. We cannot have A < mK/L 2 ,
since r could then not be positive for any value of 0. For a given E, the
turning points are solutions of the equation
<F W = 7 + & = ^ (3 " 239)
The solutions are
J^ mK \(?nK\ 2 , 2mffl 1/2
f! _ L2 + L\ L2 / + L2 J
1 mK_ \(mK\ 2 , 2mE] 112
r 2 ~ V L\ L* ) + L 2 J
(3-240)
Comparing Eq. (3-238) with Eq. (3-240), we see that the value of A in
terms of the energy and angular momentum is given by
A2== ^ + 2gtf. (3 _ 241)
The orbit is now determined in terms of the initial conditions.
An ellipse is defined as the curve traced by a particle moving so that
the sum of its distances from two fixed points F, F' is constant.* The
* For a more detailed treatment of conic sections, see W. F. Osgood and
W. C. Graustein, Plane and Solid Analytic Geometry. New York: Macmillan,
1938. (Chapters 6, 7, 8, 10.)
3-1 4] INVERSE SQUARE LAW FORCE
129
Fig. 3-38. Geometry of the ellipse.
points F, F' are called the foci of the ellipse. Using the notation indicated
in Fig. 3-38, we have
r' + r = 2a, (3-242)
where a is half the largest diameter (major axis) of the ellipse. In terms
of polar coordinates with center at the focus F and with the negative z-axis
through the focus F', the cosine law gives
r' 2 = r 2 + 4a 2 e 2 + 4rae cos 0, (3-243)
where ae is the distance from the center of the ellipse to the focus, e is
called the eccentricity of the ellipse. If e = 0, the foci coincide and the
ellipse is a circle. As € -* 1, the ellipse degenerates into a parabola or
straight line segment, depending on whether the focus F' recedes to in-
finity or remains a finite distance from F . Substituting r' from Eq. (3-242)
in Eq. (3-243), we find
r= l + ecosV < 3 " 244 )
This is the equation of an ellipse in polar coordinates with the origin at one
focus. If b is half the smallest diameter (minor axis), we have, from
Fig. 3-38,
b = a(l- e 2 ) 1 ' 2 . (3-245)
The area of the ellipse can be obtained in a straightforward way by in-
tegration :
S = wab. (3-246)
A hyperbola is denned as the curve traced by a particle moving so that
the difference of its distances from two fixed foci F, F' is constant (Fig.
3-39). A hyperbola has two branches defined by
r' — r = 2a (+ branch),
(3-247)
r — r = —2a (— branch).
130 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
D
+branch
-branch
Fig. 3-39. Geometry of the hyper-
bola.
Fig. 3-40. Geometry of the para-
bola.
We shall call the branch which encircles F the + branch (left branch in the
figure), and the branch which avoids F, the — branch (right branch in the
figure). Equation (3-243) holds also for the hyperbola, but the eccen-
tricity e is now greater than one. The equation of the hyperbola becomes
in polar coordinates:
«(c 2 - 1)
±1 + CCOS0
(3-248)
(The + sign refers to the + branch, the — sign to the — branch.) The
asymptotes of the hyperbola (dotted lines in Fig. 3-39) make an angle a
with the axis through the foci, where a is the value of for which r is infinite :
cos a = ±
1
(3-249)
A parabola is the curve traced by a particle moving so that its distance
from a fixed line D (the directrix) equals its distance from a fixed focus F.
From Fig. 3-40, we have
a
1 + cos ' (3-250)
r =
where a is the distance from the focus F to the directrix D.
We can write the equations for all three conic sections in the standard
form
- = B + A cos 0,
r
(3-251)
3_14 1 INVERSE SQUARE LAW FORCE 131
where A is positive, and B and A are given as follows:
B > A, ellipse,
B = «(1 - «*) ' ^ = a( i - e2) J (3-252)
B = A, parabola,
* = £• ^ = £; (3-253)
< B < A, hyperbola, + branch,
—A<B<0, hyperbola, — branch,
*= -«(</_ D' A = a(e2 *_ iy (3-255)
The case B < —A cannot occur, since r would then not be positive for any
value of 6. If we allow an arbitrary orientation of the curve with respect
to the z-axis, then Eq. (3-251) becomes
- = B + A cos (0 - 6 ), (3-256)
where 6 is the angle between the z-axis and the line from the origin to the
perihelion (point of closest approach of the curve to the origin). It will
be noted that in all cases
€ = -\Bj- (3-267)
For an ellipse or hyperbola,
B
a =
A* - £2
(3-258)
Equation (3-237) for the orbit of a particle under an inverse square law
force has the form of Eq. (3-256) for a conic section, with [if we use
Eq. (3-241)]
-(■
(3-259)
* + «!)"■.
132 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
The eccentricity of the orbit, by Eq. (3-257), is
--( i + 2 S) m - < 3 - 26o)
For an attractive force (K < 0), the orbit is an ellipse, parabola, or hyper-
bola, depending on whether E < 0, E = 0, or E > 0; if a hyperbola, it is
the + branch. For a repulsive force (K > 0), we must have E > 0, and
the orbit can only be the — branch of a hyperbola. These results agree
with our preliminary qualitative discussion. For elliptic and hyperbolic
orbits, the semimajor axis a is given by
a =
K
2E
(3-261)
It is curious that this relation does not involve the eccentricity or the angu-
lar momentum; the energy E depends only on the semimajor axis a, and
vice versa. Equations (3-260) and (3-261) may be obtained directly from
Eq. (3-239) for the turning points of the r motion. If we solve this equa-
tion for r, we obtain the turning points
ri - 2 = 2E ±
[(0+£eT- ^
The maximum and minimum radii for an ellipse are
r li2 = a(l ± e), (3-263)
and the minimum radius for a hyperbola is
ri = a(e =F 1), (3-264)
where the upper sign is for the + branch and the lower sign for the —
branch. Comparing Eqs. (3-263) and (3-264) with Eq. (3-262), we can
read off the values of a and e. Thus if we know that the path is an ellipse
or hyperbola, we can find the size and shape from Eq. (3-239), which fol-
lows from the simple energy method of treatment, without going through
the exact solution of the equation for the orbit. This is a useful point to
remember.
3-15 Elliptic orbits. The Kepler problem. Early in the seventeenth
century, before Newton's discovery of the laws of motion, Kepler an-
nounced the following three laws describing the motion of the planets, de-
duced from the extensive and accurate observations of planetary motions
by Tycho Brahe:
3-1 5J ELLIPTIC ORBITS. THE KEPLER PROBLEM 133
(1) The planets move in ellipses with the sun at one focus.
(2) Areas swept out by the radius vector from the sun to a planet in
equal times are equal.
(3) The square of the period of revolution is proportional to the cube
of the semimajor axis.
The second law is expressed by our Eq. (3-226), and is a consequence of
the conservation of angular momentum; it shows that the force acting on
the planet is a central force. The first law follows, as we have shown,
from the fact that the force is inversely proportional to the square of the
distance. The third law follows from the fact that the gravitational force
is proportional to the mass of the planet, as we now show.
In the case of an elliptical orbit, we can find the period of the motion
from Eqs. (3-227) and (3-246):
2m . 1m 2/1 2sl/2 (ir 2 K 2 m\ 112
r = ^ r 7rab = - r ^(l-er' 2 ={^^) , (3-265)
or, using Eq. (3-261),
T a = 4*V|£
(3-266)
In the case of a small body of mass m moving under the gravitational at-
traction [Eq. (3-230)] of a large body of mass M, this becomes
2 47T 2 s
T = MG a - (3 ~ 267 )
The coefficient of a 3 is now a constant for all planets, in agreement with
Kepler's third law. Equation (3-267) allows us to "weigh" the sun, if we
know the value of G, by measuring the period and major axis of any plane-
tary orbit. This has already been worked out in Chapter 1, Problem 9,
for a circular orbit. Equation (3-267) now shows that the result applies
also to elliptical orbits if the semimajor axis is substituted for the radius.
We have shown that Kepler's laws follow from Newton's laws of motion
and the law of gravitation. The converse problem, to deduce the law of
force from Kepler's laws and the law of motion, is an easier problem, and a
very important one historically, for it was in this way that Newton de-
duced the law of gravitation. We expect that the motions of the planets
should show slight deviations from Kepler's laws, in view of the fact that
the central force problem which was solved in the last section represents an
idealization of the actual physical problem. In the first place, as pointed
out in Section 3-13, we have assumed that the sun is stationary, whereas
actually it must wobble slightly due to the attraction of the planets going
134 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
around it. This effect is very small, even in the case of the largest planets,
and can be corrected for by the methods explained later in Section 4-7.
In the second place, a given planet, say the earth, is acted on by the gravi-
tational pull of the other planets, as well as by the sun. Since the masses
of even the heaviest planets are only a few percent of the mass of the sun,
this will produce small but measurable deviations from Kepler's laws. The
expected deviations can be calculated, and they agree with the very pre-
cise astronomical observations. In fact, the planets Neptune and Pluto
were discovered as a result of their effects on the orbits of the other planets.
Observations of the planet Uranus for about sixty years after its discovery
in 1781 showed unexplained deviations from the predicted orbit, even after
corrections were made for the gravitational effects of the other known
planets. By a careful and elaborate mathematical analysis of the data,
Adams and Leverrier were able to show that the deviations could be ac-
counted for by assuming an unknown planet beyond Uranus, and they cal-
culated the position of the unknown planet. The planet Neptune was
promptly discovered in the predicted place.
The orbits of the comets, which are occasionally observed to move in
around the sun and out again, are, at least in some cases, very elongated
ellipses. It is not at present known whether any of the comets come from
beyond the solar system, in which case they would, at least initially, have
parabolic or hyperbolic orbits. Even those comets whose orbits are known
to be elliptical have rather irregular periods due to the perturbing gravita-
tional pull of the larger planets near which they occasionally pass. Be-
tween close encounters with the larger planets, a comet will follow fairly
closely a path given by Eq. (3-256), but during each such encounter, its
motion will be disturbed, so that afterwards the constants A, B, and
will have values different from those before the encounter.
As noted in Section 3-13, we expect in general that the bounded orbits
arising from an attractive central force F(r) will not be closed (Fig. 3-34).
Closed orbits (except for circular orbits) arise only where the period of
radial oscillations is equal to, or is an exact rational multiple of, the period
of revolution. Only for certain special forms of the function F(r), of which
the inverse square law is one, will the orbits be closed. Any change in the
inverse square law, either a change in the exponent of r or an addition to
F{r) of a term not inversely proportional to r 2 , will be expected to lead to
orbits that are not closed. However, if the change is very small, then the
orbits ought to be approximately elliptical. The period of revolution will
then be only slightly greater or slightly less than the period of radial oscilla-
tions, and the orbit will be approximately an ellipse whose major axis
rotates slowly about the center of force. As a matter of fact, a slow pre-
cession of the major axis of the orbit of the planet Mercury has been ob-
served, with an angular velocity of 41 seconds of arc per century, over and
3-16] HYPERBOLIC ORBITS. THE RUTHERFORD PROBLEM
135
above the perturbations accounted for by the gravitational effects of the
other planets. It was once thought that this could be accounted for by
the gravitational effect of dust in the solar system, but it can be shown
that the amount of dust is far too small to account for the effect. It is
now fairly certain that the effect is due to slight corrections to Newton's
theory of planetary motion required by the theory of relativity.*
The problem of the motion of electrons around the nucleus of an atom
would be the same as that of the motion of planets around the sun, if
Newtonian mechanics were applicable. Actually, the motion of electrons
must be calculated from the laws of quantum mechanics. Before the dis-
covery of quantum mechanics, Bohr was able to give a fair account of the
behavior of atoms by assuming that the electrons revolve in orbits given
by Newtonian mechanics. Bohr's theory is still useful as a rough picture
of atomic structure, f
3-16 Hyperbolic orbits. The Rutherford problem. Scattering cross
section. The hyperbolic orbits are of interest in connection with the mo-
tion of particles around the sun which may come from or escape to outer
space, and also in connection with the collisions of two charged particles.
If a light particle of charge q t encounters a heavy particle of charge q 2 at
rest, the light particle will follow a hyperbolic trajectory past the heavy
particle, according to the results obtained in Section 3-14. In the case
of collisions of atomic particles, the region in which the trajectory bends
from one asymptote to the other is very small (a few angstrom units or
less), and what is observed is the deflection angle © = ir — 2a (Fig. 3-41)
between the paths of the incident particle before and after the collision.
Figure 3-41 is drawn for the case of a repelling center of force at F, but the
figure may also be taken to represent the case of an attracting center at
F'. By Eqs. (3-249) and (3-260),
tan | = cot a = (e » - i)-i/» = Qgg) 1 ' 2 • (3-268)
Let the particle have an initial speed v , and let it be traveling in such a
direction that, if undeflected, it would pass a distance s from the center of
force (F). The distance s is called the impact parameter for the collision.
We can readily compute the energy and angular momentum in terms of
*
A. Einstein and L. Infeld, The Evolution of Physics. New York: Simon and
Schuster, 1938. (Page 253.) For a mathematical discussion, see R. C. Tolman,
Relativity, Thermodynamics, and Cosmology. Oxford: Oxford University Press,
1934. (Section 83.)
t M. Born, Atomic Physics, tr. by John Dougall. New York: Stechert, 1936.
(Chapter 5.)
136
MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
Fig. 3-41. A hyperbolic orbit.
the speed and impact parameter:
E = \mo%, (3-269)
L = mv s. (3-270)
Substituting in Eq. (3-268), we have for the scattering angle 0:
tan-^ = J^- 2 - (3-271)
2 msv
If a light particle of charge q t collides with a heavy particle of charge q 2 ,
this is, by Eq. (3-231),
tan ® = JM?L (3-272)
2 msv
In a typical scattering experiment, a stream of charged particles may be
shot in a definite direction through a thin foil. Many of the particles
emerge from the foil in a different direction, after being deflected or scat-
tered through an angle 6 by a collision with a particle within the foil.
To put Eq. (3-272) in a form in which it can be compared with experiment,
we must eliminate the impact parameter s, which cannot be determined
experimentally. In the experiment, the fraction of incident particles scat-
tered through various angles is observed. It is customary to express
the results in terms of a cross section denned as follows. If N incident
particles strike a thin foil containing n scattering centers per unit area, the
average number dN of particles scattered through an angle between and
3-16]
HYPERBOLIC ORBITS. THE RUTHERFORD PROBLEM
137
Fig. 3-42. Cross section for scattering.
+ d® is given in terms of the cross section da by the formula
dN
N
= nda.
(3-273)
da is called the cross section for scattering through an angle between and
© + d®, and can be thought of as the effective area surrounding the scat-
tering center which the incident particle must hit in order to be scattered
through an angle between and © + d®. For if there is a "target area"
da around each scattering center, then the total target area in a unit area
is n da. If N particles strike one unit area, the average number striking
the target area is Nn da, and this, according to Eq. (3-273), is just dN, the
number of particles scattered through an angle between © and © + d®.
Now consider an incident particle approaching a scattering center F
as in Figs. 3-41 and 3-42. If the impact parameter is between s and
s + ds, the particle will be scattered through an angle between and
© + d®, where is given by Eq. (3-272), and d® is given by the differen-
tial of Eq. (3-272) :
2cos2(0/2)
d®
Igigzl
ms 2 Vo
(3-274)
The area of the ring around F of inner radius s, outer radius s + ds, at
which the incident particle must be aimed in order to be scattered through
an angle between and + d®, is
da = 27rs ds. (3-275)
Substituting for s from Eq. (3-272), and for ds from Eq. (3-274) (omitting
138 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
the negative sign), we obtain
\2m»g/ sinM©/2)
This formula can be compared with da determined experimentally as given
by Eq. (3-273). Formula (3-276) was deduced by Rutherford and used in
interpreting his experiments on the scattering of alpha particles by thin
metal foils. He was able to show that the formula agrees with his experi-
ments with gi = 2e (charge on alpha particle),* and q 2 = Ze (charge on
atomic nucleus), so long as the perihelion distance (a + ae in Fig. 3-41) is
larger than about 10 -12 cm, which shows that the positive charge on the
atom must be concentrated within a region of radius less than 10~~ 12 cm.
This was the origin of the nuclear theory of the atom. The perihelion
distance can be computed from formula (3-262) or by using the conserva-
tion laws for energy and angular momentum, and is given by
3i«2
2E
\ mqilzj
(3-277)
The smallest perihelion distance for incident particles of a given energy
occurs when L = (s = 0), and has the value
r lmi „ = ^- (3-278)
Hence if there is a deviation from Coulomb's law of force when the alpha
particle grazes or penetrates the nucleus, it should show up first as a devia-
tion from Rutherford's law [Eq. (3-276)] at large angles of deflection @, and
should show up when the energy E is large enough so that
E > ^ , (3-279)
where r is the radius of the nucleus. The earliest measurements of nuclear
radii were made in this way by Rutherford, and turn out to be of the order
of 10 -12 cm.
The above calculation of the cross section is strictly correct only when
the alpha particle impinges on a nucleus much heavier than itself, since the
scattering center is assumed to remain fixed. This restriction can be re-
moved by methods to be discussed in Section 4-8. Alpha particles also
collide with electrons, but the electron is so light that it cannot appreciably
deflect the alpha particle. The collision of an alpha particle with a nucleus
* Here e stands for the magnitude of the electronic charge.
3-17] MOTION OF A PARTICLE IN AN ELECTROMAGNETIC FIELD 139
should really be treated by the methods of quantum mechanics. The con-
cept of a definite trajectory with a definite impact parameter s is no longer
valid in quantum mechanics. The concept of cross section is still valid in
quantum mechanics, however, as it should be, since it is denned in terms of
experimentally determined quantities. The final result for the scattering
cross section turns out the same as our formula (3-276).* It is a fortunate
coincidence in the history of physics that classical mechanics gives the right
answer to this problem.
3-17 Motion of a particle in an electromagnetic field. The laws deter-
mining the electric and magnetic fields due to various arrangements of
electric charges and currents are the subject matter of electromagnetic
theory. The determination of the motions of charged particles under given
electric and magnetic forces is a problem in mechanics. The electric force
on a particle of charge q located at a point r is
F = qE(r), (3-280)
where E(r) is the electric field intensity at the point r. The electric field
intensity may be a function of time as well as of position in space. The
force exerted by a magnetic field on a charged particle at a point r depends
on the velocity v of the particle, and is given in terms of the magnetic
induction B(r) by the equation.f
F = 2 VX B(r), (3-281)
where c = 3 X 10 10 cm/sec is the velocity of light, and all quantities are
in gaussian units, i.e., q is in electrostatic units, B in electromagnetic units
(gauss), and v and F are in cgs units. In mks units, the equation reads
F = qv X B(r). (3-282)
Equation (3-280) holds for either gaussian or mks units. We shall base
our discussion on Eq. (3-281) (gaussian units), but the results are readily
transcribed into mks units by omitting c wherever it occurs. The total
electromagnetic force acting on a particle due to an electric field intensity E
and a magnetic induction B is
F = qE + I v X B. (3-283)
* D. Bohm, Quantum Theory. New York: Prentice-Hall, 1951. (Page 537).
t G. P. Harnwell, Principles of Electricity and Electromagnetism, 2nd ed. New
York: McGraw-Hill, 1949. (Page 302.)
140 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
If an electric charge moves near the north pole of a magnet, the magnet
will exert a force on the charge given by Eq. (3-281) ; and by Newton's
third law the charge should exert an equal and opposite force on the
magnet. This is indeed found to be the case, at least when the velocity
of the particle is small compared with the speed of light, if the magnetic
field due to the moving charge is calculated and the force on the magnet
computed. However, since the magnetic induction B is directed radially
away from the pole, and the force F is perpendicular to B, the forces on
the charge and on the pole are not directed along the line joining them,
as in the case of a central force. Newton's third law is sometimes stated
in the "strong" form in which action and reaction are not only equal and
opposite, but are directed along the line joining the interacting particles.
For magnetic forces, the law holds only in the "weak" form in which nothing
is said about the directions of the two forces except that they are opposite.
This is true not only of the forces between magnets and moving charges,
but also of the magnetic forces exerted by moving charges on one another.
If the magnetic field is constant in time, then the electric field intensity
can be shown to satisfy the equation
V X E = 0. (3-284)
The proof of this statement belongs to electromagnetic theory and need
not concern us here.* We note, however, that this implies that for static
electric and magnetic fields, the electric force on a charged particle is con-
servative. We can therefore define an electric potential
*(r) = - pE-dr, (3-285)
Jr ,
such that
E = —V*. (3-286)
Since E is the force per unit charge, <j> will be the potential energy per unit
charge associated with the electric force:
F(r) = g*(r). (3-287)
Furthermore, since the magnetic force is perpendicular to the velocity, it
can do no work on a charged particle. Consequently, the law of con-
servation of energy holds for a particle in a static electromagnetic field:
T + q<t> = E, (3-288)
where E is a constant.
* Harnwell, op. cit. (Page 340.)
3-17] MOTION OF A PARTICLE IN AN ELECTROMAGNETIC FIELD 141
A great variety of problems of practical and theoretical interest arise
involving the motion of charged particles in electric and magnetic fields.
In general, special methods of attack must be devised for each type of
problem. We shall discuss two special problems which are of interest both
for the results obtained and for the methods of obtaining those results.
We first consider the motion of a particle of mass m, charge q, in a uni-
form constant magnetic field. Let the z-axis be chosen in the direction of
the field, so that
B(r, t) = Bk, (3-289)
where B is a constant. The equations of motion are then, by Eq. (3-281),
mx = — y, my = — —x, mz = 0. (3-290)
c c
According to the last equation, the z-component of velocity is constant,
and we shall consider the case when v z = 0, and the motion is entirely in
the a;?/-plane. The first two equations are not hard to solve, but we can
avoid solving them directly by making use of the energy integral, which in
this case reads
imv 2 = E. (3-291)
The force is given by:
F = ^vxk, (3-292)
F = ^- (3-293)
The force, and consequently the acceleration, is therefore of constant
magnitude and perpendicular to the velocity. A particle moving with
constant speed v and constant acceleration a perpendicular to its direction
of motion moves in a circle of radius r given by Eq. (3-80) :
2
a = rd 2 = — = — • (3-294)
r m
,2 v< F
We substitute for F from Eq. (3-293) and solve for r:
The product Br is therefore proportional to the momentum and inversely
proportional to the charge.
142 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
This result has many practical applications. If a cloud chamber is placed in a
uniform magnetic field, one can measure the momentum of a charged particle by
measuring the radius of curvature of its track. The same principle is used in a
beta-ray spectrometer to measure the momentum of a fast electron by the curva-
ture of its path in a magnetic field. In a mass spectrometer, a particle is ac-
celerated through a known difference of electric potential, so that, by Eq. (3-288),
its kinetic energy is
imv 2 = q(<j>o - «i). (3-296)
It is then passed through a uniform magnetic field B. If q is known, and r, B,
(<po — 4>i) are measured, we can eliminate v between Eqs. (3-295) and (3-296),
and solve for the mass:
- - qBV (3-297)
2 C 2(4> - 0i)
There are many variations of this basic idea. The historic experiments of
J. J. Thomson which demonstrated the existence of the electron were essentially
of this type, and by them Thomson succeeded in showing that the path traveled
by a cathode ray is that which would be followed by a stream of charged particles,
all with the same ratio q/m. In a cyclotron, charged particles travel in circles in
a uniform magnetic field, and receive increments in energy twice per revolution
by passing through an alternating electric field. The radius r of the circles there-
fore increases, according to Eq. (3-295), until a maximum radius is reached, at
which radius the particles emerge in a beam of definite energy determined by
Eq. (3-295). The frequency v of the alternating electric field must be the same
as the frequency v of revolution of the particles, which is given by
v = 2jm>. (3-298)
Combining this equation with Eq. (3-295), we have
- - qB (3-299)
2Tr?nc
Thus if B is constant, v is independent of r, and this is the fundamental principle
on which the operation of the cyclotron is based.* In the betatron, electrons
travel in circles, and the magnetic field within the circle is made to increase.
Since B is changing with time, V X E is no longer zero; the changing magnetic
flux induces a voltage around the circle such that a net amount of work is done
on the electrons by the electric field as they travel around the circle. The betatron
is so designed that the increase of B at the electron orbit is proportional to the
increase of mv, so that r remains constant.
* According to the theory of relativity, the mass of a particle increases with
velocity at velocities near the speed of light, and consequently the cyclotron
cannot accelerate particles to such speeds unless v is reduced or B is increased as
the particle velocity increases. [It turns out that Eq. (3-295) still holds in
relativity theory.]
3-17] MOTION OF A PARTICLE IN AN ELECTROMAGNETIC FIELD 143
Finally, we consider a particle of mass m, charge q, moving in a uniform
constant electric field intensity E and a uniform constant magnetic induc-
tion B. Again let the z-axis be chosen in the direction of B, and let the
y-axis be chosen so that E is parallel to the t/z-plane:
B = Bk, E = E v j + EJk, (3-300)
where B, E y , E z are constants. The equations of motion, by Eq. (3-283),
are
.. qB .
mx= =c y ' (3-301)
my = —2-& + q E y> (3-302)
mz = qE z . (3-303)
The z-component of the motion is uniformly accelerated:
* = 20 + z t + | & t 2 . (3-304)
To solve the x and y equations, we differentiate Eq. (3-301) and substitute
in Eq. (3-302) in order to eliminate y.
m 2 c ... qB . , _,
Tr x = - , x + &»■ (3-305)
qB'"~ c
By making the substitutions
qB
U) =
mc
(3-306)
qE v
* = ^> (3-307)
we can write Eq. (3-305) in the form
-^p+u> £ x = aw. (3-308)
This equation has the same form as the equation for a harmonic oscillator
with angular frequency w subject to a constant applied "force" aw, except
that x appears in place of the coordinate. The corresponding oscillator
problem was considered in Chapter 2, Problem 33. The solution in this
case will be
* = £ + A x cos (uf + e x ), (3-309)
where A x and 6 X are arbitrary constants to be determined. By eliminating x
144 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
from Eqs. (3-301) and (3-302), in a similar way, we obtain a solution for y:
y = A y cos (at + 6 V ). (3-310)
We get x and y by integrating Eqs. (3-309) and (3-310):
* = C x + - + ^ sin (co* + 8 X ), (3-311)
CO CO
y = C y + 4+ sin (co< + «,)• (3-312)
CO
Now a difficulty arises, for we have six constants A x , A v , 6 X , y , C x , and C y
to be determined, and only four initial values Xq, yo, xq, yo to determine
them. The trouble is that we obtained the solutions (3-311) and (3-312) by
differentiating the original equations, and differentiating an equation may
introduce new solutions that do not satisfy the original equation. Con-
sider, for example, the very simple equation
x = 3.
Differentiating, we get
x = 0,
whose solution is
x = C.
Now only for one particular value of the constant C will this satisfy the
original equation. Let us substitute Eqs. (3-311) and (3-312) or, equiva-
lents, Eqs. (3-309) and (3-310) into the original Eqs. (3-301) and (3-302),
using Eqs. (3-306) and (3-307):
- 2? A x sin (cot + 6 X ) = ^- A y cos (co< + 0„), (3-313)
c c
- 2® Ay sin (at + e v ) = - 2? A x cos (at + 6 X ). (3-314)
c c
These two equations will hold only if A X9 A y> B x , and B y are chosen so that
sin (cof + X ) = —cos (at + 8 V ), (3-316)
cos (at + 6 X ) = sin (co< + 6 y ). (3-317)
The latter two equations are satisfied if
*»=«. + £• (3-318)
3-17] MOTION OF A PARTICLE IN AN ELECTBOMAGNETIC FIELD 145
Fig. 3-43. Orbits in the xy-plane of a charged particle subject to a magnetic
field in the z-direction and an electric field in the ^-direction.
Let us set
A x
e x
Ay = coA,
8,= e + %-
Then Eqs. (3-311) and (3-312) become
x = C x + A sin (cot +■ 6) +
V = Cy + A cos (cot + 6).
at
CO
(3-319)
(3-320)
(3-321)
(3-322)
(3-323)
There are now only four constants, A, 6, C x , C y , to be determined by the
initial values x , yo, x 0) y . The z-motion is, of course, given by Eq. (3-304).
If E y = 0, the xy-motion is in a circle of radius A with angular velocity w
about the point (C x , C y ) ; this is the motion considered in the previous
example. The effect of E y is to add to this uniform circular motion a
uniform translation in the x-direction! The resulting path in the :n/-plane
will be a cycloid having loops, cusps, or ripples, depending on the initial
conditions and on the magnitude of E v (Fig. 3-43). This problem is of
interest in connection with the design of magnetrons.
146 motion op particle in two or three dimensions [chap. 3
Problems
1. Prove, on the basis of the geometric definitions of the operations of vector
algebra, the following equations. In many cases a diagram will suffice, (a) Eq.
(3-7), (b) Eq. (3-17), (c) Eq. (3-26), (d) Eq. (3-27), *(e) Eq. (3-35).
2. Prove, on the basis of the algebraic definitions of the operations of vector
algebra in terms of components, the following equations: (a) Eq. (3^8), (b) Eq.
(3-17), (c) Eq. (3-27), (d) Eq. (3-34), (e) Eq. (3-35).
3. Derive Eq. (3-32) by direct calculation, using Eq. (3-10) to represent A
and B, and making use of Eqs. (3-25) to (3-31).
4. (a) Prove that A • (B X C) is the volume of the parallelepiped whose edges
are A, B, C with positive or negative sign according to whether a right-hand screw
rotated from A toward B would advance along C in the positive or negative direc-
tion. A, B, C are any three vectors not lying in a single plane, (b) Use this
result to prove Eq. (3-34) geometrically. Verify that the right and left members
of Eq. (3-34) are equal in sign as well as in magnitude.
5. Prove the following inequalities. Give a geometric and an algebraic proof
(in terms of components) for each:
(a) |A + B| < |A| + |B|.
(b) |A • B| < |A| |B|.
(0 |A X B| < |A| |B|.
6. (a) Obtain a formula analogous to Eq. (3-40) for the magnitude of the
sum of three forces Fi, F2, F3, in terms of Fi, F2, F3, and the angles 812, 023, 031
between pairs of forces. [Use the suggestions following Eq. (3-40).]
(b) Obtain a formula in the same terms for the angle ai, between the total
force and the component force Fi.
7. Prove Eqs. (3-54) and (3-55) from the definition (3-52) of vector differen-
tiation.
8. Prove Eqs. (3-56) and (3-57) from the algebraic definition (3-53) of vector
differentiation.
9. Give suitable definitions, analogous to Eqs. (3-52) and (3-53), for the
integral of a vector function A(<) with respect to a scalar t:
f
J 4,
A(«) dt
Write a set of equations like Eqs. (3-54)-(3-57) expressing the algebraic proper-
ties you would expect such an integral to have. Prove that on the basis of either
definition
d -f
dtJo
A(t) dt = A(t).
10. A 45° isosceles right triangle ABC has a hypotenuse AB of length 4a. A
particle is acted on by a force attracting it toward a point O on the hypotenuse a
distance a from the point A. The force is equal in magnitude to k/r 2 , where r is
PROBLEMS 147
the distance of the particle from the point 0. Calculate the work done by this
force when the particle moves from A to C to B along the two legs of the triangle.
Make the calculation by both methods, that based on Eq. (3-61) and that based
on Eq. (3-63).
11. (a) A particle in the xy-pl&ne is attracted toward the origin by a force
F = k/y, inversely proportional to its distance from the z-axis. Calculate the
work done by the force when the particle moves from the point x = 0, y = a to
the point x = 2a, y = along a path which follows the sides of a rectangle
consisting of a segment parallel to the z-axis from x = 0, y = a to x = 2a,
y = a, and a vertical segment from the latter point to the z-axis. (b) Calculate
the work done by the same force when the particle moves along an ellipse of
semiaxes a, 2a. [Hint: Set x = 2a sin 0, y = a cos 0.]
12. (a) Find the components of d 3 i/dt 3 in spherical coordinates, (b) Find
the components of d 2 A/dt 2 in cylindrical polar coordinates, where the vector A
is a function of t and is located at a moving point.
*13. (a) Plane parabolic coordinates /, h are defined in terms of cartesian
coordinates x, y by the equations
*-/-*» V = 2W 1 ' 2 ,
where / and h are never negative. Find / and h in terms of x and y. Let unit
vectors f , h be defined in the directions of increasing / and h respectively. That
is, f is a unit vector in the direction in which a point would move if its /-coordinate
increases slightly while its ^-coordinate remains constant. Show that f and h are
perpendicular at every point. [Hint: f = (i dx + j dy)[(dx) 2 + (dy) 2 ]~ 1/2 , when
df > 0, dh = 0. Why?]
(b) Show that f and h are functions of /, h, and find their derivatives with
respect to / and h. Show that r = / 1/2 (/ + h) 1/2 f + h ll2 <J + h) V2 h. Find
the components of velocity and acceleration in parabolic coordinates.
14. A particle moves along the parabola
y = 4/o — 4/oz,
where /o is a constant. Its speed v is constant. Find its velocity and acceleration
components in rectangular and in polar coordinates. Show that the equation of
the parabola in polar coordinates is
r cos - = /o.
What is the equation of this parabola in parabolic coordinates (Problem 13)?
15. A particle moves with varying speed along an arbitrary curve lying in the
zy-plane. The position of the particle is to be specified by the distance s the par-
ticle has traveled along the curve from some fixed point on the curve. Let r(s)
be a unit vector tangent to the curve at the point s in the direction of increasing s.
Show that
dr v
148 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
where v(s) is a unit vector normal to the curve at the point s, and r(s) is the radius
of curvature at the point s, defined as the distance from the curve to the point of
intersection of two nearby normals* Hence derive the following formulas for the
velocity and acceleration of the particle:
.2
v = st, a = st -\ v .
r
16. Using the properties of the vector symbol V, derive the vector identities:
curl (curl A) = grad (div A) — V 2 A,
u grad v = grad (uv) — v grad u.
Then write out the ^-components of each side of these equations and prove by
direct calculation that they are equal in each case. (One must be very careful,
in using the first identity in curvilinear coordinates, to take proper account of the
dependence of the unit vectors on the coordinates.)
17. Calculate curl A in cylindrical coordinates.
18. Give a suitable definition of the angular momentum of a particle about an
axis in space. Taking the specified axis as the z-axis, express the angular momen-
tum in terms of cylindrical coordinates. If the force acting on the particle has
cylindrical components F z , F„, F v , prove that the time rate of change of angular
momentum about the z-axis is equal to the torque about that axis.
19. A moving particle of mass m is located by spherical coordinates r(t), 0(t),
<p(t). The force acting on it has spherical components F r , Ft, F v . Calculate the
spherical components of the angular momentum vector and of the torque vector
about the origin, and verify by direct calculation that the equation
dL
follows from Newton's equation of motion.
20. Solve for the next term beyond those given in Eqs. (3-177) and (3-178).
21. A projectile is to be fired from the origin in the zz-plane (z-axis vertical)
with muzzle velocity wo to hit a target at the point x = xo, z = 0. (a) Neglect-
ing air resistance, find the correct angle of elevation of the gun. Show that, in
general, there are two such angles unless the target is at or beyond the maximum
range, (b) Find the first-order correction to the angle of elevation due to air
resistance.
22. A projectile is fired from the origin with initial velocity vo = (v xo , v yo , v zo ).
The wind velocity is v K = wj. Solve the equations of motion (3-180) for x, y, z
as functions of t. Find the point x\, y\ at which the projectile will return to the
horizontal plane, keeping only first-order terms in 6. Show that if air resistance
and wind velocity are neglected in aiming the gun, air resistance alone will cause
the projectile to fall short of its target a fraction 4bv 20 /3mg of the target distance,
* W. F. Osgood, Introduction to the Calculus. New York: Macmillan, 1937,
259.
PROBLEMS 149
and that the wind causes an additional miss in the ^-coordinate of amount
2bwv 2 J(mg 2 ).
23. Determine which of the following forces are conservative, and find the
potential energy for those which are:
(a) F x = 6abz 3 y — 20bx 3 y 2 , F y = 6abxz 3 — 10bx 4 y,
F z = 18abxz 2 y.
(b) F x = 18abyz 3 — 20bx 3 y 2 , F y = 18ate 3 — \Qbx*y,
F z = 6abxyz 2 .
(c) F = iF.(x) + jF y (y) + W.(«).
24. Determine the potential energy for any of the following forces which are
conservative:
(a) F x = 2ax(z 3 + y 3 ), F y = 2ay(z 3 + y 3 ) + 3ay 2 (x 2 + y 2 ),
F z = 3az 2 (x 2 + y 2 ).
(b)
F„ = ap 2 cos <p, F v =
= ap 2 sin <p, F z = 2az 2 .
(c)
F r = — 2ar sin cos <p,
F v = ar sin sin ip.
F$ = — ar cos 6 cos <p,
25. A particle is attracted toward the 2-axis by a force proportional to the
square of its distance from the xy-plane and inversely proportional to its distance
from the «-axis. Add an additional perpendicular force in such a way as to
make the total force conservative, and find the potential energy. Be sure to
write expressions for the forces and potential energy which are dimensionally
consistent.
26. Find the components of force for the following potential-energy functions:
(a) V = axy 2 z 3 .
(b) V = ikr 2 .
(c) V = ik*x 2 + ik v y 2 + ik z z 2 .
27. Find the force on the electron in the hydrogen molecule ion for which the
potential is
2 2
e e
where r\ is the distance from the electron to the point y = z = 0, x = — a,
and t2 is the distance from the electron to the point y = z = 0, x = a.
28. Show that F = nF(r) (where n is a unit vector directed away from the
origin) is a conservative force by showing by direct calculation that the integral
I
r 2
F-dr
along any path between ri and T2 depends only on r\ and r-2. [Hint: Express F
and dr in spherical coordinates.]
150 MOTION OF PARTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3
29. The potential energy for an isotropic harmonic oscillator is
V = ikr 2 .
Plot the effective potential energy for the r-motion when a particle of mass m
moves with this potential energy and with angular momentum L about the
origin. Discuss the types of motion that are possible, giving as complete a descrip-
tion as is possible without carrying out the solution. Find the frequency of
revolution for circular motion and the frequency of small radial oscillations
about this circular motion. Hence describe the nature of the orbits which differ
slightly from circular orbits.
30. Find the frequency of small radial oscillations about steady circular mo-
tion for the effective potential given by Eq. (3-232) for an attractive inverse
square law force, and show that it is equal to the frequency of revolution.
31. Find r{t), 8(t) for the orbit of the particle in Problem 29. Compare with
the orbits found in Section 3-10 for the three-dimensional harmonic oscillator.
32. A particle of mass m moves under the action of a central force whose poten-
tial is
V(r) = Kr 4 , K > 0.
For what energy and angular momentum will the orbit be a circle of radius a
about the origin? What is the period of this circular motion? If the particle is
slightly disturbed from this circular motion, what will be the period of small
radial oscillations about r = o?
33. According to Yukawa's theory of nuclear forces, the attractive force be-
tween a neutron and a proton has the potential
Ke~ ar
V(r) = — — , K < 0.
r
(a) Find the force, and compare it with an inverse square law of force, (b) Dis-
cuss the types of motion which can occur if a particle of mass m moves under
such a force, (c) Discuss how the motions will be expected to differ from the
corresponding types of motion for an inverse square law of force, (d) Find L
and E for motion in a circle of radius a. (e) Find the period of circular motion
and the period of small radial oscillations, (f) Show that the nearly circular
orbits are almost closed when a is very small.
34. (a) Discuss by the method of the effective potential the types of motion to
be expected for an attractive central force inversely proportional to the cube of
the radius:
(b) Find the ranges of energy and angular momentum for each type of motion.
(c) Solve the orbital equation (3-222), and show that the solution is one of the
forms:
PROBLEMS 151
- = a cos me - e )), (i)
r
- = A cosh [0(0 - 6o)), (2)
r
1
r
1
r
1 1 ±09
= A sinh LQ(<? - 0o)L (3)
= A(d - 0o), (4)
= - e ±p °. (5)
- r tq
(d) For what values of L and E does each of the above types of motion occur?
Express the constants A and /? in terms of E and L for each case, (e) Sketch a
typical orbit of each type.
35. (a) Discuss the types of motion that can occur for a central force
™ --*+%-
Assume that K > 0, and consider both signs for K'.
(b) Solve the orbital equation, and show that the bounded orbits have the form
(if L 2 > —mK')
1 + e cos ad
(c) Show that this is a precessing ellipse, determine the angular velocity of pre-
cession, and state whether the precession is in the same or in the opposite direc-
tion to the orbital angular velocity.
36. A comet is observed a distance of 1.00 X 10 8 km from the sun, travel-
ing toward the sun with a velocity of 51.6 km per second at an angle of 45°
with the radius from the sun. Work out an equation for the orbit of the comet
in polar coordinates with origin at the sun and z-axis through the observed
position of the comet. (The mass of the sun is 2.00 X 10 30 kgm.)
37. It will be shown in Chapter 6 (Problem 5) that the effect of a uniform dis-
tribution of dust of density p about the sun is to add to the gravitational attrac-
tion of the sun on a planet, of mass m an additional attractive central force
F' = —mhr,
where
(a) If the mass of the sun is M , find the angular velocity of revolution of the
planet in a circular orbit of radius ro, and find the angular frequency of small
radial oscillations. Hence show that if F' is much less than the attraction due
152 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
to the sun, a nearly circular orbit will be approximately an ellipse whose major
axis precesses slowly with angular velocity
(b) Does the axis precess in the same or in the opposite direction to the orbital
angular velocity? Look up M and the radius of the orbit of Mercury, and cal-
culate the density of dust required to cause a precession of 41 seconds of arc
per century.
38. It can be shown (Chapter 6, Problems 15 and 19) that the correction to
the potential energy of a mass m in the earth's gravitational field, due to the
oblate shape of the earth, is approximately, in spherical coordinates, relative
to the polar axis of the earth,
, rimMGR 2 2 m
V = =-= (1—3 cos 0),
where M is the mass of the earth and 2R, 2R(l — ij) are the equatorial and
polar diameters of the earth. Calculate the rate of precession of the perigee
(point of closest approach) of an earth satellite moving in a nearly circular
orbit in the equatorial plane. Look up the mass of the earth and the equatorial
and polar diameters, and estimate the rate of precession in degrees per revolution
for a satellite 400 miles above the earth.
*39. Calculate the torque on an earth satellite due to the oblateness potential
energy correction given in Problem 38. A satellite moves in a circular orbit of
radius r whose plane is inclined so that its normal makes an angle a with the
polar axis. Assume that the orbit is very little affected in one revolution, and
calculate the average torque during a revolution. Show that the effect of such a
torque is to make the normal to the orbit precess in a cone of half angle a about
the polar axis, and find a formula for the rate of precession in degrees per revolu-
tion. Calculate the rate for a satellite 400 miles above the earth, using suitable
values for M , i\, and R.
40. (a) A satellite is to be launched from the surface of the earth. Assume
the earth is a sphere of radius R, and neglect friction with the atmosphere. The
satellite is to be launched at an angle a with the vertical, with a velocity v o, so
as to coast without power until its velocity is horizontal at an altitude hi above
the earth's surface. A horizontal thrust is then applied by the last stage rocket
so as to add an additional velocity Awi to the velocity of the satellite. The final
orbit is to be an ellipse with perigee hi (point of closest approach) and apogee
hi (point farthest away) measured from the earth's surface. Find the required
initial velocity vq and additional velocity Ai>i, in terms of R, a, hi, h%, and g,
the acceleration of gravity at the earth's surface.
(b) Write a formula for the change Shi in perigee height due to a small error
5/3 in the final thrust direction, to order (5/3) 2 .
41. Two planets move in the same plane in circles of radii n, r^ about the
sun. A space probe is to be launched from planet 1 with velocity v\ relative
to the planet, so as to reach the orbit of planet 2. (The velocity vi is the relative
PROBLEMS 153
velocity after the probe has escaped from the gravitational field of the planet.)
Show that f i is a minimum for an elliptical orbit whose perihelion and aphelion
are r\ and r^. In that case, find v\, and the relative velocity V2 between the
space probe and planet 2 if the probe arrives at radius T2 at the proper time to
intercept planet 2. Express your results in terms of n, r%, and the length of
the year Y\ of planet 1. Look up the appropriate values of r\ and r^, and esti-
mate v\ for trips to Venus and Mars from the earth.
42. A rocket is in an elliptical orbit around the earth, perigee r\, apogee r^,
measured from the center of the earth. At a certain point in its orbit, its engine
is fired for a short time so as to give a velocity increment Aw in order to put
the rocket on an orbit which escapes from the earth with a final velocity wo
relative to the earth. (Neglect any effects due to the sun and moon.) Show
that Ay is a minimum if the thrust is applied at perigee, parallel to the orbital
velocity. Find Aw in that case in terms of the elliptical orbit parameters e, a,
the acceleration g at a distance R from the earth's center, and the final velocity
wo- Can you explain physically why Aw is smaller for larger e?
43. A satellite moves around the earth in an orbit which passes across the
poles. The time at which it crosses each parallel of latitude is measured so that
the function 0(t) is known. Show how to find the perigee, the semimajor axis,
and the eccentricity of its orbit in terms of 0(0, and the value of g at the surface
of the earth. Assume the earth is a sphere of radius R.
44. It can be shown that the orbit given by the special theory of relativity for
a particle of mass m moving under a potential energy V(r) is the same as the orbit
which the particle would follow according to Newtonian mechanics if the poten-
tial energy were
where E is the energy (kinetic plus potential), and c is the speed of light. Discuss
the nature of the orbits for an inverse square law of force according to the theory
of relativity. Show by comparing the orbital angular velocity with the frequency
of radial oscillations for nearly circular motion that the nearly circular orbits,
when the relativistic correction is small, are precessing ellipses, and calculate
the angular velocity of precession.
45. A particle of mass m moves in an elliptical orbit of major axis 2a, eccentric-
ity e, in such a way that the radius to the particle from the center of the ellipse
sweeps out area at a constant rate
dS
dt ~ L '
and with period r independent of a and e. (a) Write out the equation of the ellipse
in polar coordinates with origin at the center of the ellipse, (b) Show that the
force on the particle is a central force, and find F(r) in terms of m, t.
46. A rocket moves with initial velocity wo toward the moon of mass M ,
radius ro. Find the cross section a for striking the moon. Take the moon to be
at rest, and neglect all other bodies.
154 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3
47. Show that for a repulsive central force inversely proportional to the cube
of the radius,
K
r z
FW--J. K>0,
the orbits are of the form (1) given in Problem 34, and express /8 in terms of K, E,
L, and the mass m of the incident particle. Show that the cross section for
scattering through an angle between and + d® for a particle subject to
this force is
&r-»LE *-e d®.
mvl 2 (2tt - 0) 2
48. A velocity selector for a beam of charged particles of mass m, charge e,
is to be designed to select particles of a particular velocity vo- The velocity
selector utilizes a uniform electric field E in the a;-direction and a uniform
magnetic field B in the y-direction. The beam emerges from a narrow slit along
the y-axis and travels in the 2-direction. After passing through the crossed
fields for a distance I, the beam passes through a second slit parallel to the first
and also in the yz-plane.
(a) If a particle leaves the origin with a velocity »o at a small angle with the
2-axis, find the point at which it arrives at the plane z = I. Assume that the
initial angle is small enough so that second-order terms in the angle may be
neglected.
(b) What is the best choice of E, B in order that as large a fraction as possible
of the particles with velocity vo arrive at the second slit, while particles of other
velocities miss the slit as far as possible?
(c) If the slit width is h, what is the maximum velocity deviation Sv from
vo for which a particle moving initially along the 2-axis can pass through the
second slit? Assume that E, B have the values chosen in part (b).
49. A particle of charge q in a cylindrical magnetron moves in a uniform mag-
netic field
B = Bk,
and an electric field, directed radially outward or inward from a central wire
along the 2-axis,
E = -h,
P
where p is the distance from the 2-axis, and h is a unit vector directed radially
outward from the 2-axis. The constants a and B may be either positive or
negative, (a) Set up the equations of motion in cylindrical coordinates, (b)
Show that the quantity
2. . qB 2 „
mp <p + ~ p = K
is a constant of the motion, (c) Using this result, give a qualitative discussion,
based on the energy integral, of the types of motion that can occur. Consider all
cases, including all values of a, B, K, and E. (c) Under what conditions can
circular motion about the axis occur? (d) What is the frequency of small radial
oscillations about this circular motion?
CHAPTER 4
THE MOTION OF A SYSTEM OF PARTICLES
4-1 Conservation of linear momentum. Center of mass. We consider
in this chapter the behavior of mechanical systems containing two or more
particles acted upon by internal forces exerted by the particles upon one
another, and by external forces exerted upon particles of the system by
agents not belonging to the system. We assume the particles to be point
masses each specified by its position (a;, y, z) in space, like the single par-
ticle whose motion was studied in the preceding chapter.
Let the system we are studying contain N particles, and let them be
numbered 1, 2, . . . , N. The masses of the particles we designate by
mi, m 2 , . . . , mjv. The total force acting on the fcth particle will be the sum
of the internal forces exerted on particle k by all the other (AT — 1) parti-
cles in the system, plus any external force which may be applied to particle
k. Let the sum of the internal forces on particle k be Ff., and let the total
external force on particle k be F|. Then the equation of motion of the
fcth particle will be
m£ k = ¥i + F$, k = 1, 2, . . . , N. (4-1)
The N equations obtained by letting fc in Eqs. (4-1) run over the num-
bers 1, . . . , 2V are the equations of motion of our system. Since each of
these 2V equations is itself a vector equation, we have in general a set of 32V
simultaneous second-order differential equations to be solved. The solu-
tion will be a set of functions rjt(<) specifying the motion of each particle in
the system. The solution will depend on 62V "arbitrary" constants speci-
fying the initial position and velocity of each particle. The problem of
solving the set of equations (4-1) is very difficult, except in certain special
cases, and no general methods are available for attacking the 2V-body
problem, even in the case where the forces between the bodies are central
forces. The two-body problem can often be solved, as we shall see, and
some general theorems are available when the internal forces satisfy certain
conditions.
If P* = WftVfc is the linear momentum of the kth particle, we can write
Eqs. (4-1) in the form
^ = n + H, k=l,...,N. (4-2)
155
156 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
Summing the right and left sides of these equations over all the particles,
we have
Ef = |E^=EFl+EFi (4-3)
Jfe=l Jfc=l Jb=l k=l
We designate by P the total linear momentum of the particles, and by F
the total external force:
N N
p = E p* = E m * v *> t 4-4 )
jb=i &=i
N
I
F = £ **• (4-5)
We now make the assumption, to be justified below, that the sum of the
internal forces acting on all the particles is zero:
E Pi = 0. (4-6)
k=l
When Eqs. (4-4), (4-5), and (4-6) are substituted in Eq. (4-3), it becomes
dP
dt
= F. (4-7)
This is the momentum theorem for a system of particles. It states that the
time rate of change of the total linear momentum is equal to the total
external force. An immediate corollary is the conservation theorem for
linear momentum, which states that the total momentum P is constant
when no external forces act.
We now try to justify the assumption (4-6). Our first proof is based
on Newton's third law. We assume that the force Fj. acting on particle k
due to all the other particles can be represented as a sum of separate forces
due to each of the other particles:
Fi = X) FL*, . (4-8)
where F\^ k is the force on particle k due to particle 1. According to New-
ton's third law, the force exerted by particle I on particle k is equal and
opposite to that exerted by k on I:
FJLi = -Fi-*. (4-9)
4-1] CONSERVATION OF LINEAR MOMENTUM. CENTER OP MASS 157
Equation (4-9) expresses Newton's third law in what we may call the weak
form; that is, it says that the forces are equal and opposite, but does not
imply that the forces act along the line joining the two particles. If we
now consider the sum in Eq. (4-6), we have
E F ^=SE F ^ (4-10)
fc=l k=l l^k
The sum on the right is over all forces acting between all pairs of particles
in the system. Since for each pair of particles k, I, two forces F i k _ >l and
F!_»j; appear in the total sum, and by Eq. (4-9) the sum of each such pair
is zero, the total sum on the right in Eq. (4-10) vanishes, and Eq. (4-6)
is proved.
Thus Newton's third law, in the form (4-9), is sufficient to guarantee the
conservation of linear momentum for a system of particles, and it was for
this purpose that the law was introduced. The law of conservation of
momentum has, however, a more general validity than Newton's third
law, as we shall see later. We can derive assumption (4-6) on the basis of
a somewhat weaker assumption than Newton's third law. We do not need
to assume that the particles interact in pairs. We assume only that the
internal forces are such that they would do no net work if every particle in
the system should be displaced the same small distance 5r from its position
at any particular instant. An imagined motion of all the particles in the
system is called a virtual displacement. The motion described, in which
every particle moves the same small distance 5r, is called a small virtual
translation of the system. We assume, then, that in any small virtual
translation 5r of the entire system, the internal forces would do no net
work. From the point of view of the general idea of conservation of energy,
this assumption amounts to little more than assuming that space is homo-
geneous. If we move the system to a slightly different position in space
without otherwise disturbing it, the internal state of the system should be
unaffected, hence in particular the distribution of various kinds of energy
within it should remain the same and no net work can have been done by
the internal forces. Let us use this idea to prove Eq. (4-6). The work
done by the force FJ. in a small virtual translation St is
8W k = Fl-Sr. (4-11)
The total work done by all the internal forces is
8W = X) SW " = « r '( E F *J ' ( 4 ~ 12 )
fc=i \fc=i /
158 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
where we have factored out 5r from the sum, since it is the same for all
particles. Assuming that dW = 0, we have
Sr-[ V fA = 0. (4-13)
<JH- ft
Since Eq. (4-13) must hold for any 5r, Eq. (4-6) follows.
We can put Eq. (4-7) in an illuminating form by introducing the con-
cept of center of mass of the system of particles. The vector R which
locates the center of mass is defined by the equation
MR = X) m ^k, (4-14)
where M is the total mass:
N
k=i
The coordinates of the center of mass are given by the components of
Eq. (4-14) :
X = mYj mkXk > Y = m ^ mkn ' Z = M ^ m * *" ^ 4-16 ^
k=l fc=l fc=l
The total momentum defined by Eq. (4—1) is, in terms of the center of mass,
N
so that Eq. (4-7) can be written
MR = F. (4-18)
This equation has the same form as the equation of motion of a particle of
mass M acted on by a force F. We thus have the important theorem that
[when Eq. (4-6) holds] the center of mass of a system of particles moves like
a single particle, whose mass is the total mass of the system, acted on by a force
equal to the total external force acting on the system.
4-2 Conservation of angular momentum. Let us calculate the time rate
of change of the total angular momentum of a system of N particles rela-
tive to a point Q not necessarily fixed in space. The vector angular mo-
mentum of particle k about a point Q, not necessarily the origin, is to be
defined according to Eq. (3-142) :
L*q = m k (r k — t q ) X (ft - i Q ), (4-19)
4-2] CONSERVATION OF ANGULAR MOMENTUM 159
where r e is the position vector of the point Q, and (r k — i Q ) is the vector
from Q to particle k. Note that in place of the velocity ik we have
written the velocity (ik — iq) relative to the point Q as origin, so that
LkQ is the angular momentum of m^ calculated as if Q were a fixed
origin. This is the most useful way to define the angular momentum about
a moving point Q. Taking the cross product of (rj, — iq) with the equa-
tion of motion (4-2) for particle k, as in the derivation of Eq. (3-144), we
obtain
(r* - r Q ) X ^ = (r» - i Q ) X F% + (r* - r Q ) X F|. (4-20)
We now differentiate Eq. (4-19) :
^p = (r* - r Q ) X ^ + m k (ik - f Q ) X (f* - f Q ) - m*(r ft - r ) X r Q .
(4-21)
The second term on the right vanishes. Therefore, by Eq. (4-20),
^f = (r* - r Q ) x ¥% + (r k - t q ) x r k - m k (t k - r Q ) x t Q .
(4-22)
The total angular momentum and total external torque about the point Q
are defined as follows:
N e = £ (r* - r ) X FJ. (4-24)
N.
z
Summed over all particles, Eq. (4-22) becomes, if we use Eq. (4-14),
AT
E
k=i
^f = N Q + E (** - '«) X F* - ^( R - r «) x f Q- (4-26)
The last term will vanish if the acceleration of the point Q is zero or is
along the line joining Q with the center of mass. We shall restrict the
discussion to moments about a point Q satisfying this condition:
(R - r Q ) X t Q = 0. (4-26)
The most important applications will be to cases where Q is at rest, or
where Q is the center of mass. If we also assume that the total internal
160 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
torque vanishes: N
£ (r* - t Q ) X Ft = 0, (4-27)
then Eq. (4-25) becomes
^ = Ng. (4-28)
This is the angular momentum theorem for a system of particles. An
immediate corollary is the conservation theorem for angular momentum,
which states that the total angular momentum of a system of particles is
constant if there is no external torque on the system.
In order to prove Eq. (4-27) from Newton's third law, we need to assume
a stronger version of the law than that needed in the preceding section,
namely, that the force Ff.^ is not only equal and opposite to Fj'_, fc , but
that these forces act along the line joining the two particles; that is, the
two particles can only attract or repel each other. We shall assume, as
in the previous section, that Fj. is the sum of forces due to each of the
other particles :
E (r fc - r e ) x Fit = £ £ < r * - r «) x F <-*
k=l *=1 l*k
= E £ [(r* - v Q ) x FL* + (r, - r ) x P^,].
fc=1 l=1 (4-29)
In the second step, the sum of torques has been rearranged as a sum of
pairs of torques due to pairs of forces which, according to Newton's third
law, are equal and opposite [Eq. (4-9)], so that
£ (r* - r Q ) X F| = f; £ [(r* - r ) - (r, - r«)] X FL*
= E E (r* - r,) X FL*. (4-30)
k=l 1=1
The vector (r k — t{) has the direction of the line joining particle I with
particle k. If F}_> k acts along this line, the cross product in Eq. (4-30)
vanishes. Hence if we assume Newton's third law in the strong form, then
assumption (4-27) can be proved.
Alternatively, by assuming that no net work is done by the internal
forces in a small virtual rotation about any axis through the point Q, we
can show that the component of total internal torque in any direction is
zero, and hence justify Eq. (4-27).
4r-2] CONSERVATION OF ANGULAR MOMENTUM 161
rxF
L + dL
Fig. 4-1. Motion of a simple gyroscope.
As an application of Eq. (4-28), we consider the action of a gyroscope
or top. A gyroscope is a rigid system of particles symmetrical about an
axis and rotating about that axis. The reader can convince himself that
when the gyroscope is rotating about a fixed axis, the angular momentum
vector of the gyroscope about a point Q on the axis of rotation is directed
along the axis of rotation, as in Fig. 4-1. The symmetry about the axis
guarantees that any component of the angular momentum Lj of particle
k that is perpendicular to the axis will be compensated by an equal and
opposite component due to the diametrically opposite particle. Let us
choose the point Q where the gyroscope axis rests on its support. If now
a force F is applied downward on the gyroscope axis (e.g., the force of
gravity), the torque (r x F) due to F will be directed perpendicular to r
and to L, as shown in Fig. 4-1. By Eq. (4-28) the vector dL/dt is in the
same direction, as shown in the figure, and the vector L tends to precess
around the figure in a cone under the action of the force F. Now the
statement that L is directed along the gyroscope axis is strictly true only
if the gyroscope is simply rotating about its axis. If the gyroscope axis
itself is changing its direction, then this latter motion will contribute an
additional component of angular momentum. If, however, the gyroscope
is spinning very rapidly, then the component of angular momentum along
its axis will be much greater than the component due to the motion of the
axis, and L will be very nearly parallel to the gyroscope axis. Therefore
the gyroscope axis must also precess around the vertical, remaining essen-
tially parallel to L. A careful analysis of the off-axis components of L
shows that, if the gyroscope axis is initially stationary in a certain direction
and is released, it will wobble slightly down and up as it precesses around
the vertical. This will be shown in Chapter 11. The gyroscope does not
"resist any change in its direction, " as is sometimes asserted, for the rate
162 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
of change in its angular momentum is always equal to the applied torque,
just as the rate of change of linear momentum is always equal to the ap-
plied force. We can make the gyroscope turn in any direction we please
by applying the appropriate torque. The importance of the gyroscope as
a directional stabilizer arises from the fact that the angular momentum
vector L remains constant when no torque is applied. The changes in
direction of a well-made gyroscope are small because the applied torques
are small and L is very large, so that a small dL gives no appreciable change
in direction. Furthermore, a gyroscope only changes direction while a
torque is applied; if it shifts slightly due to occasional small frictional
torques in its mountings, it stops shifting when the torque stops. A large
nonrotating mass, if mounted like a gyroscope, would acquire only small
angular velocities due to frictional torques, but once set in motion by a
small torque, it would continue to rotate, and the change in position might
eventually become large.
4-3 Conservation of energy. In many cases, the total force acting on
any particle in a system of particles depends only on the positions of the
particles in the system :
F t = F*(ri, r 2 , . . . , ly), k = 1, 2, . . . , N. (4-31)
The external force FJ, for example, might depend on the position r* of
particle k, and the internal force Fj. might depend on the positions of the
other particles relative to particle k. It may be that a potential function
y( T it r 2> • • • > t n) exists such that
F - dV F - dV F - dV h- 1 N
Fkx --te k ' Fk «--w k ' Fk *-~df k ' k - 1> ---> N -
(4-32)
Conditions to be satisfied by the force functions Ffc(ri, . . . , r^) in order
for a potential V to exist can be worked out, analogous to the condition
(3-189) for a single particle. The result is rather unwieldy and of little
practical importance, and we omit this development here. If a potential
energy exists, we can derive a conservation of energy theorem as follows.
By Eq. (4-32), the equations of motion of the fcth particle are
(4-33)
Multiplying Eqs. (4-33) by v kx , v ky , Vkz, respectively, and adding, we have
for each k:
d ft m „ 2 \ _1_ dV dXk _1_ dV d W° _j_ d V dz k n h , jj ,, oa\
dVicx
_s>Z,
dv ky
dvkz
_ dV
m "-dT =
dx k '
mk -w =
— — ,
tyk'
"*-* =
dZk
4-3] CONSERVATION OF ENERGY 163
This is to be summed over all values of k :
d V n 2 N i v^ IdV dzjc , dV dy k . dV dzA _ ,. __ N
j t ^^) + Z\^- kW + ei kW + er kW ) = o. (4-35)
The second term in Eq. (4-35) is dV/dt:
$r = ^ (w dx± dV fa av dzj\
dt £l W dt d Vk dt dz k dt/' { axy)
and the first term is the time derivative of the total kinetic energy
AT
T = X) i m ^- ( 4 - 37 )
Consequently, Eq. (4-35) can be written
j t (T + V) = 0. (4-38)
Hence we again have a conservation of energy theorem,
T + V = E, (4-39)
where E is constant. If the internal forces are derivable from a potential-
energy function V, as in Eq. (4-32), but the external forces are not, the
energy theorem will be
j t (T + V) = JT ¥%-v k . (4-40)
Suppose the internal force acting on any particle A; can be regarded as
the sum of forces due to each of the other particles, where the force Ff^j.
on k due to I depends only on the relative position (tk — rj) of particle k
with respect to particle I:
Ft = X) Fi-^bfob - r,). (4-41)
It may be that the vector function F|_ >Jfc (rj ; — rj) is such that we can define
a potential-energy function
V k i(T k i) = - jf" FL*(r t ,)-<frw f (4-42)
where
m = n — Ti- (4-43)
This will be true if F\^, k is a conservative force in the sense of Chapter 3,
164 THE MOTION OF A SYSTEM OP PARTICLES [CHAP. 4
that is, if
curl FL* = 0, (4-44)
where the derivatives are with respect to x k i, yu, zm- The gravitational and
electrostatic forces between pairs of particles are examples of conservative
forces. If Ff_j. is conservative, so that Vu can be denned, then*
V i -dVki .dV„ dV k i
*'-* * dx kl J dy kl dz k i
~ l dx k 3 dy k * dz k ^*° ;
If Newton's third law (weak form) holds, then
oxjci dy k i dz k i
= _i*Z*i_ j^lM -^-IM. (4-46)
dxi dyi dzi
Thus V k i will also serve as the potential-energy function for the force Fj^.
We can now define the total internal potential energy V 1 for the system of
particles as the sum of Vu over all the pairs of particles :
F'(ri, . . . , r N ) = j^ *fj V k i(v k - r,). (4-47)
k =i 1=1
It follows from Eqs. (4-41), (4-45), and (4-46), that the internal forces are
given by
F£=-i^-J^-k^, k=l,...,N. (4-48)
dx k dy k dz k
In particular, if the forces between pairs of particles are central forces, the
potential energy V k i(r k i) for each pair of particles depends only on the
distance r k i between them, and is given by Eq. (3-200) ; the internal forces
of the system are then conservative, and Eq. (4—48) holds. The energy
theorem (4-40) will be valid for such a system of particles. If the external
forces are also conservative, their potential energy can be added to V 1 , and
the total energy is constant.
If there is internal friction, as is often the case, the internal frictional
forces depend on the relative velocities of the particles, and the conserva-
tion law of potential plus kinetic energy no longer holds.
* Note that V(x k i) = V(x kh y kh z u ) = V(x k — x h y k — yi, z k — z t ), so that
dV/dx k = dV/dx H = — dV/dxi, etc.
4 ~ 4 J CRITIQUE OP THE CONSERVATION LAWS
165
4-4 Critique of the conservation laws. We may divide the phenomena
to which the laws of mechanics have been applied into three major classes.
The motions of celestial bodies— stars, satellites, planets— are described
with extremely great precision by the laws of classical mechanics. It was
in this field that the theory had many of its important early successes.
The motions of the bodies in the solar system can be predicted with great
accuracy for periods of thousands of years. The theory of relativity pre-
dicts a few slight deviations from the classically predicted motion, but these
are too small to be observed except in the case of the orbit of Mercury,
where relativity and observation agree in showing a slow precession of the
axis of the elliptical orbit around the sun at an angular velocity of about
0.01 degree per century.
The motion of terrestrial bodies of macroscopic and microscopic size
constitutes the second major division of phenomena. Motions in this class
are properly described by Newtonian mechanics, without any significant
corrections, but the laws of force are usually very complicated, and often
not precisely known, so that the beautifully precise calculations of celestial
mechanics cannot be duplicated here.
The third class of phenomena is the motion of "atomic" particles:
molecules, atoms, electrons, nuclei, protons, neutrons, etc. Early attempts
to describe the motions of such particles were based on classical mechanics,
and many phenomena in this class can be understood and predicted on this
basis. However, the finer details of the behavior of atomic particles can
only be. properly described in terms of quantum mechanics and, for high
velocities, relativistic quantum mechanics must be introduced. We might
add a fourth class of phenomena, having to do with the intrinsic structure
of the elementary particles themselves (protons, neutrons, electrons, etc.).
Even quantum mechanics fails to describe such phenomena correctly, and
physics is now struggling to produce a new theory which will describe this
class of phenomena.
The conservation law for linear momentum holds for systems of celestial
bodies as well as for bodies of macroscopic and microscopic size. The
gravitational a'nd mechanical forces acting between such bodies satisfy
Newton's third law, at least to a high degree of precision. Linear momen-
tum is also conserved in most interactions of particles of atomic size, except
when high velocities or rapid accelerations are involved. The electrostatic
forces between electric charges at rest satisfy Newton's third law, but when
the charges are in motion, their electric fields propagate with the velocity
of light, so that if two charges are in rapid relative motion, the forces be-
tween them may not at any instant be exactly equal and opposite. If a
fast electron moves past a stationary proton, the proton "sees" the electron
always a little behind its actual position at any instant, and the force on
the proton is determined, not by where the electron is, but by where it was
166 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
a moment earlier. When electric charges accelerate, they may emit electro-
magnetic radiation and lose momentum in so doing. It turns out that the
law of conservation of momentum can be preserved also in such cases, but
only by associating momentum with the electromagnetic field as well as
with moving particles. Such a redefinition of momentum goes beyond the
original limits of Newtonian mechanics.
Celestial bodies and bodies of macroscopic or microscopic size are ob-
viously not really particles, since they have a structure which for many
purposes is not adequately represented by merely giving to the body three
position coordinates x, y, z. Nevertheless, the motion of such bodies, in
problems where their structure can be neglected, is correctly represented
by the law of motion of a single particle,
mx = F. (4-49)
This is often justified by regarding the macroscopic body as a system
of smaller particles satisfying Newton's third law. For such a system,
the linear momentum theorem holds, and can be written in the form of
Eq. (4-18), which has the same form as Eq. (4-49). This is a very con-
venient way of justifying the application of Eq. (4-49) to bodies of macro-
scopic or astronomical size, provided our conscience is not troubled by the
fact that according to modern ideas it does not make sense. If the particles
of which the larger body is composed are taken as atoms and molecules,
then in the first place Newton's third law does not invariably hold for such
particles, and in the second place we should apply quantum mechanics, not
classical mechanics, to their motion. The momentum theorem (4-18) can
be derived for bodies made up of atoms by using the laws of electrodynam-
ics and quantum mechanics, but this lies outside the scope of Newtonian
mechanics. Hence, for the present, we must take the law of motion (4-49),
as applied to macroscopic and astronomical bodies, as a fundamental
postulate in itself, whose justification is based on experimental grounds
or on the results of deeper theories. The theorems proved in Section 4-1
show that this postulate gives a consistent theory of mechanics in the
sense that if, from bodies satisfying this postulate, we construct a com-
posite body, the latter body will also satisfy the postulate.
The law of conservation of angular momentum, as formulated in Sec-
tion 4-2 for a system of particles, holds for systems of celestial bodies
(regarded as particles) and for systems of bodies of macroscopic size when-
ever effects due to rotation of the individual bodies can be neglected.
When rotations of the individual bodies enter into the motion, then a
conservation law for angular momentum still holds, provided we include
the angular momentum associated with such rotations; the bodies are then
no longer regarded as particles of the simple type considered in the pre-
ceding sections whose motions are completely described simply by specify-
4-4] CRITIQUE OF THE CONSERVATION LAWS 167
ing the function r(t) for each particle. The total angular momentum of
the solar system is very nearly constant, even if the sun, planets, and
satellites are regarded as simple particles whose rotations can be neglected.
Tidal forces, however, convert some rotational angular momentum into
orbital angular momentum of the planets and satellites, and so rotational
angular momentum must be included if the law of conservation of angular
momentum is to hold precisely. Some change in angular momentum occurs
due to friction with interplanetary dust and rocks, but the effect is too
small to be observed, and could in any case be included by adding the
angular momentum of the interplanetary matter to the total.
The law of conservation of total angular momentum, including rotation,
of astronomical and terrestrial bodies can be justified by regarding each
body as a system of smaller particles whose mutual forces satisfy Newton's
third law (strong form). The argument of Section 4-2 then gives the law
of conservation of total angular momentum, the rotational angular momen-
tum of a body appearing as ordinary orbital angular momentum (r x p) of
the particles of which it is composed. This argument is subject to the same
criticism as applied above to the case of linear momentum. If the "par-
ticles" of which a body is composed are atoms and molecules, then Newton's
third law does not always hold, particularly in its strong form; moreover,
the laws of quantum mechanics apply to such particles; and in addition
atoms and molecules also possess rotational angular momentum which
must be taken into account. Even the elementary particles — electrons,
protons, neutrons, etc. — possess an intrinsic angular momentum which
is not associated with their orbital motion. This angular momentum is
called spin angular momentum from its analogy with the intrinsic angular
momentum of rotation of a macroscopic body, and must be included if the
total is to satisfy a conservation law. Thus we never arrive at the ideal
simple particle of Newtonian mechanics, described by its position r(t) alone.
We are left with the choice of accepting the conservation law of angular
momentum as a basic postulate, or appealing for its justification to theories
which go beyond classical mechanics.
The gravitational forces acting between astronomical bodies are con-
servative, so that the principle of conservation of mechanical energy holds
very accurately in astronomy. In principle, there is a small loss of mechan-
ical energy in the solar system due to friction with interplanetary dust and
rocks, but the effect is too small to produce any observable effects on plane-
tary motion, even with the high precision with which astronomical events
are predicted and observed. There is also a very gradual but measurable
loss of rotational energy of planets and satellites due to tidal friction. For
terrestrial bodies of macroscopic or microscopic size, friction usually plays
an important part, and only in certain special cases where friction may be
neglected can the principle of conservation of energy in the form (4-39)
168 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
or even (4-40) be applied. However, it was discovered by Joule that we
can associate energy with heat in such a way that the law of conservation
of energy of a system of bodies still applies to the total kinetic plus poten-
tial plus heat energy. If we regard a body as composed of atoms and mole-
cules, its heat energy turns out to be kinetic and potential energy of ran-
dom motion of its atoms and molecules. The electromagnetic forces on
moving charged particles are not conservative, and an electromagnetic
energy must be associated with the electromagnetic field in order to pre-
serve the conservation law of energy. Such extensions of the concept of
energy to include heat and electromagnetic energy are, of course, outside
the domain of mechanics. When the definition of energy is suitably ex-
tended to include not only kinetic energy, but energy associated with the
electromagnetic fields and any other force fields which may act, then a
law of conservation of energy holds quite generally, in classical, relativistic,
and quantum physics.
The conservation laws of energy, momentum, and angular momentum
are the cornerstones of present-day physics, being generally valid in all
physical theories. It seems at present an idle exercise to attempt to prove
them for material bodies within the framework of classical mechanics by
appealing to an outmoded picture of matter as made up of simple New-
tonian particles exerting central forces upon one another. The conserva-
tion laws are in a sense not laws at all, but postulates which we insist must
hold in any physical theory. If, for example, for moving charged particles,
we find that the total energy, defined as (T + V), is not constant, we do
not abandon the law, but change its meaning by redefining energy to in-
clude electromagnetic energy in such a way as to preserve the law. We
prefer always to look for quantities which are conserved, and agree to
apply the names "total energy," "total momentum," "total angular mo-
mentum" only to such quantities. The conservation of these quantities
is then not a physical fact, but a consequence of our determination to de-
fine them in this way. It is, of course, a statement of physical fact, which
may or may not be true, to assert that such definitions of energy, momen-
tum, and angular momentum can always be found. This assertion, has
so far been true; a deeper justification will be suggested at the end of
Section 9-6.
4-5 Rockets, conveyor belts, and planets. There are many problems
that can be solved by appropriate applications of the conservation laws
of linear momentum, angular momentum, and energy. In solving such
problems, it is necessary to decide which conservation laws are appropriate.
The conservation laws of linear and angular momentum or, rather, the
theorems (4-7) and (4-28) of which they are corollaries, are always appli-
cable to any physical system provided all external forces and torques are
4-5] ROCKETS, CONVEYOR BELTS, AND PLANETS 169
taken into account, and application of one or the other is appropriate
whenever the external forces or torques are known. The law of conserva-
tion of kinetic plus potential energy is applicable only when there is no
conversion of mechanical energy into other forms of energy. We cannot
use the law of conservation of energy when there is friction, for example,
unless there is a way to determine the amount of heat energy produced.
The conservation laws of energy, momentum, and angular momentum
refer always to a definite fixed system of particles. In applying the con-
servation laws, care must be taken to decide just how much is included in
the system to which they are to be applied, and to include all the energy
and momentum of this system in writing down the equations. One may
choose the system arbitrarily, including and excluding whatever particles
may be convenient, but if any forces act from outside the system on parti-
cles in the system, these must be taken into account.
A typical problem in which the law of conservation of linear momentum
is applicable is the conveyor belt problem. Material is dropped continu-
ously from a hopper onto a moving belt, and it is required to find the force
F required to keep the belt moving at constant velocity v (Fig. 4-2). Let
the rate at which mass is dropped on the belt be dm/dt. If to is the mass
of material on the belt, and M is the mass of the belt (which really does
not figure in the problem), the total momentum of the system, belt plus
material on the belt and in the hopper, is
P = (to + M)v. (4-50)
We assume that the hopper is at rest; otherwise the momentum of the
hopper and its contents must be included in Eq. (4-50). The linear mo-
mentum theorem requires that
„ dP dm ,. _,.
This gives the force applied to the belt. The power supplied by the force is
This is twice the rate at which the kinetic energy is increasing, so that the
_© ® © <s>
Fig. 4-2. A conveyor belt.
170
THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
conservation theorem of mechanical energy (4-40) does not apply here.
Where is the excess half of the power going?
The equation of motion of a rocket can be obtained from the law of
conservation of momentum. Let the mass of the rocket at any given in-
stant be M , and let its speed be v relative to some fixed coordinate system.
If material is shot out of the rocket motor with an exhaust velocity u rela-
tive to the rocket, the velocity of the exhaust relative to the fixed coordi-
nate system is v + u. If an external force F also acts on the rocket, then
the linear momentum theorem reads in this case:
£ (Mv) - (v + u) ^ = F. (4-53)
The first term is the time rate of change of momentum of the rocket. The
second term represents the rate at which momentum is appearing in the
rocket exhaust, where— (dM/dt) is the rate at which matter is being ex-
hausted. The conservation law applies to a definite fixed system of par-
ticles. If we fix our attention on the rocket at any moment, we must
remember that at a time dt later this system will comprise the rocket plus
the material exhausted from the rocket during that time, and both must
be considered in computing the change in momentum. The equation can
be rewritten:
M| = u? + F. (4-54)
dt dt
The first term on the right is called the thrust of the rocket motor. Since
dM/dt is negative, the thrust is opposite in direction to the exhaust veloc-
ity. The force F may represent air resistance, or a gravitational force.
Let us solve this equation for the special case where there is no external
force:
M$ = u^- (4-55)
dt dt
We multiply by dt/M and integrate, assuming that u is constant:
v-v =-uln^- (4-56)
The change of speed in any interval of time depends only on the exhaust
velocity and on the fraction of mass exhausted during that time interval.
This result is independent of any assumption as to the rate at which mass
is exhausted.
Problems in which the law of conservation of angular momentum is use-
ful turn up frequently in astronomy. The angular momentum of the
galaxy of stars, or of the solar system, remains constant during the course
4-6] COLLISION PROBLEMS 171
of its development provided no material is ejected from the system. The
effect of lunar tides is gradually to slow down the rotation of the earth.
As the angular momentum of the rotating earth decreases, the angular
momentum of the moon must increase. The magnitude of the (orbital)
angular momentum of the moon is
L = mr 2 w, (4-57)
where m is the mass, o> is the angular velocity, and r is the radius of the
orbit of the moon. We can equate the mass times the centripetal accelera-
tion to the gravitational force, to obtain the relation
2 GMm .
mrur = — — ' (4-58)
where M is the mass of the earth. Solving this equation for w and substi-
tuting in Eq. (4-57), we obtain
L= (GMm 2 r) 1/2 . (4-59)
Therefore, as the moon's angular momentum increases, it moves farther
away from the earth. (In attempting to determine the rate of recession
of the moon by equating the change of L to the change of the earth's rota-
tional angular momentum, it would be necessary to determine how much
of the slowing down of the earth's rotation by tidal friction is due to the
moon and how much to the sun. The angular momentum of the moon
plus the rotational angular momentum of the earth is not constant because
of the tidal friction due to the sun. The total angular momentum of the
earth-moon system about the sun is very nearly constant except for the
very small effect of tides raised on the sun by the earth.)
4-6 Collision problems. Many questions concerning collisions of
particles can be answered by applying the conservation laws. Since the
conservation laws are valid also in quantum mechanics,* results obtained
with their use are valid for particles of atomic and subatomic size, as well
as for macroscopic particles. In most collision problems, the colliding
particles are moving at constant velocity, free of any force, for some time
before and after the collision, while during the collision they are under the
action of the forces which they exert on one another. If the mutual forces
during the collision satisfy Newton's third law, then the total linear mo-
mentum of the particles is the same before and after the collision. If
Newton's third law holds in the strong form, the total angular momentum
* P. A. M. Dirac, The Principles of Quantum Mechanics, 3rd ed. Oxford:
Oxford University Press, 1947. (Page 115.)
172 THE MOTION OF A SYSTEM OF PABTICLES [CHAP. 4
is conserved also. If the forces are conservative, kinetic energy is con-
served (since the potential energy before and after the collision is the same).
In any case, the conservation laws are always valid if we take into account
all the energy, momentum, and angular momentum, including that asso-
ciated with any radiation which may be emitted and including any energy
which is converted from kinetic energy into other forms, or vice versa.
We consider first a collision between two particles, 1 and 2, in which
the total kinetic energy and linear momentum are known to be conserved.
Such a collision is said to be elastic. If we designate by subscripts 1 and
2 the two particles, and by subscripts I and F the values of kinetic energy
and momentum before and after the collision respectively, the conservation
laws require
PlJ + P2/ = PlF + P2F, (4-60)
Tu + T 2I = T 1F + T 2F . (4-61)
Equation (4-61) can be rewritten in terms of the momenta and masses of
the particles:
Vu + Pf/ = Pif_ + P2F_ . (4 _g 2)
2mi 2wi2 2mi 2m
2
To specify any momentum vector p, we must specify three quantities,
which may be either its three components along any set of axes, or its mag-
nitude and direction (the latter specified perhaps by spherical angles 0, <p).
Thus Eqs. (4-60) and (4-62) represent four equations involving the ratio
of the two masses and twelve quantities required to specify the momenta
involved. If nine of these quantities are given, the equations can be solved
for the remaining four. In a typical case, we might be given the masses and
initial momenta of the two particles, and the final direction of motion of
one of the particles, say particle 1. We could then find the final momen-
tum p 2 F of particle 2, and the magnitude of the final momentum pip (or
equivalently, the energy) of particle 1. In many important cases, the mass
of one of the particles is unknown, and can be computed from Eqs. (4-60)
and (4-62) if enough is known about the momenta and energies before and
after the collision. Note that the initial conditions alone are not enough
to determine the outcome of the collision from Eqs. (4-60) and (4-62) ; we
must know something about the motion after the collision. The initial
conditions alone would determine the outcome if we could solve the equa-
tions of motion of the system.
Consider a collision of a particle of mass m 1; momentum pu, with a
particle of mass ra 2 at rest. This is a common case. (There is actually no
loss of generality in this problem, since, as we pointed out in Section 1-4
and will show in Section 7-1, if m 2 is initially moving with a uniform veloc-
4-6] COLLISION PROBLEMS 173
Pu
Fig. 4-3. Collision of particle mi with particle m 2 at rest.
ity v 2 /, Newton's laws are equally applicable in a coordinate system mov-
ing with uniform velocity v 2J , in which m 2 is initially at rest.) Let m x be
"scattered" through an angle #1; that is, let &i be the angle between its
final and its initial direction of motion (Fig. 4-3). The momentum p 2 i?
must lie in the same plane as pu and pif since there is no component of
momentum perpendicular to this plane before the collision, and there must
be none after. Let p 2 F make an angle # 2 with the direction of pu. We
write out Eq. (4-60) in components along and perpendicular to pu:
Pu = Pif cos #1 + p 2P cos t? 2 , (4-63)
= Pif sin &i — p 2F sin & 2 . (4-64)
Equation (4-62) becomes, in the present case,
2 2 2
PlI — PlF __ P2F
mi m 2
If two of the quantities
(Pif/ph, P2f/Pii, #i, #2, mi/m 2 )
(4-65)
are known, the remaining three can be found. If the masses, the initial
momentum pu, and the angle &i are known, for example, we can solve for
Pif, P2F, &2 as follows. Transposing the first term on the right to the
left side in Eqs. (4-63) and (4-64), squaring, and adding, we eliminate & 2 :
Pit + Pif — ZpuPiF cos#i = pi F . (4-66)
After substituting this in Eq. (4-65), we can solve for p 1F :
***■ = —^— cos * ± [(— ^— Y cos 2 * + m * - H 1 ' 2 ,
Pu mi + m 2 l\mi + m 2 f m x + m 2 \
(4-67;
and p 2 F can now be found from Eq. (4-66), and # 2 from Eq. (4-63)
174 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
If mi > m 2 , the quantity under the radical is zero for #1 = d m , where
# m is given by 2
cos 2 # m = 1 - ^|, < d m < f • ' (4-68)
mi
If t?i > t? m (and #i < tt), then p\ F /pu is either imaginary or negative,
neither of which is allowable physically, so that # m represents the maxi-
mum angle through which m x can be scattered. If m t 3> m 2 , this angle
is very small, as we know from experience. For t?j < & m , there are two
values of Pif/Pii, the larger corresponding to a glancing collision, the
smaller to a more nearly head-on collision; # 2 will be different for these
two cases. The case t?i = may represent either no collision at all
(p 1F = p u ) or a head-on collision. In the latter case,
Pip = tm - m 2 f p 2J . = _2m 2
Pu mi rf m 2 Pn mi + m 2
If mi = m 2 , Eqs. (4-67), (4-66), and (4-64) reduce to
^ = cos «?i, ^ = sin tfi, * a = (f - *i) • (4-70)
Pu Pu \^ '
#i now varies from # x = for no collision to t>i = ir/2 for a head-on
collision in which the entire momentum is transferred to particle 2. (Actu-
ally, &i is undefined if p 1F = 0, but j?i — >7r/2 and p iF — * as the colli-
sion approaches a head-on collision.) If m x < m 2 , all values of #i from
to 7T are possible, and give a positive value for pir/pu if the plus sign is
chosen in Eq. (4-67). The minus sign cannot be chosen, since it leads to a
negative value for pwlpu- If #i = 0i then p XF = pu\ this is the case
when there is no collision. The case &i = ir corresponds to a head-on
collision, for which
""' 0! = 7T,
(4-71)
#2 = 0.
PlF
m 2 — mi
Pu
mi + m 2
P2F
2m 2
Pu
mi + m 2
If mi is unknown, but either pu or Tu can be measured or calculated, ob-
servation of the final momentum of particle 2 (whose mass is assumed
known) is sufficient to determine wii- As an example, if Tu = pf 7 /2mi
is known, and T 2F is measured for a head-on collision, m x is given by
Eq. (4-69) or (4-71) :
m 2 J-2F L\l2F / -I
We thus determine m t to within one of two possible values. If results for
4-6] COLLISION PROBLEMS 175
a collision with another particle of different mass m 2 , or for a different
scattering angle, are known, mi is determined uniquely. Essentially this
method was used by Chadwick to establish the existence of the neutron.*
Unknown neutral particles created in a nuclear reaction were allowed to
impinge on matter containing various nuclei of known masses. The ener-
gies of two kinds of nuclei of different masses m 2 , ra 2 projected forward
by head-on collisions were measured. By writing Eq. (4-72) for both
cases, the unknown energy Tu could be eliminated, and the mass mi was
found to be practically equal to that of the proton.
We have seen that if we know the initial momenta of two colliding
particles of known masses, and the angle of scattering #1 (or t? 2 ), all other
quantities involved in the collision can be calculated from the conserva-
tion laws. To predict the angles of scattering, we must know not only
the initial momenta and the initial trajectories, but also the law of force
between the particles. An example is the collision of two particles acted
on by a central inverse square law of force, to be treated in Section 4-8.
Such predictions can be made for collisions of macroscopic or astronomical
bodies under suitable assumptions as to the law of force. For atomic par-
ticles, which obey quantum mechanics, this cannot be done, although we
can predict the probabilities of observing various angles #! (or # 2 ) for
given initial conditions; that is, we can predict cross sections. In all cases
where energy is conserved, the relationships between energies, momenta,
and angles of scattering developed above are valid except at particle veloc-
ities comparable with the velocity of light. In the latter case, Eqs. (4-60),
(4-61), (4-63), and (4-64) are still valid, but the relativistic relationships
between mass, momentum, and energy must be used, instead of Eq. (4-62).
We quote without proof the relation between mass, momentum, and
energy as given by the theory of relativity :f
where c is the speed of light, and m is the rest mass of the particle, that is,
the mass when the particle is at rest. The relativistic relations between
kinetic energy, momentum, and velocity are
T = mc 2 ( 1 - l\ > (4-74)
Wi - 0> 2 A 2 ) /
p mv = , (4-75)
\/l — (y 2 A 2 )
* J. Chadwick, Nature, 129, 312 (1932).
t P. G. Bergmann, Introduction to the Theory of Relativity. New York: Prentice-
Hall, 1946. (Chapter 6.)
176 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
which reduce to the classical relations (2-5) and (3-127), when !)«c.
Unless v is nearly equal to c, the second term on the right in Eq. (4-73)
is much smaller than the first, and this equation reduces to the classical
one. With the help of Eq. (4-73), the conservation laws can be applied to
collisions involving velocities near the speed Of light.
Atoms, molecules, and nuclei possess internal potential and kinetic
energy associated with the motion of their parts, and may absorb or re-
lease energy on collision. Such inelastic collisions between atomic particles
are said to be of the first kind, or endoergic, if kinetic energy of transla-
tional motion is absorbed, and of the second kind, or exoergic, if kinetic
energy is released in the process. It may also happen that in an atomic
or nuclear collision, the final particles after the collision are not the same
as the initial particles before collision. For example, a proton may collide
with a nucleus and be absorbed while a neutron is released and flies away.
There are a great many possible types of such processes. Two particles
may collide and stick together to form a single particle or, conversely, a
single particle may suddenly break up into two particles which fly apart.
Two particles may collide and form two other particles which fly apart.
Or three or more particles may be formed in the process and fly apart
after the collision. In all these cases, the law of conservation of momen-
tum holds, and the law of conservation of energy also if we take into
account the internal energy of the atoms and molecules. We consider here
a case in which a particle of mass mi collides with a particle of mass m 2
at rest (Fig. 4-4). Particles of masses mz and m± leave the scene of the
collision at angles # 3 and #4 with respect to the original direction of
motion of mi. Let kinetic energy Q be absorbed in the process (Q >
for an endoergic collision; Q — for an elastic collision; Q < for an
pi
Fig. 4-4. Collision of mi with TO2 at rest, resulting in the production of m%
and m,4.
4-6] COLLISION PROBLEMS 177
exoergic collision). Then, applying the conservation laws of energy and
momentum, we write
Pi = Pz cos # 3 + Pi cos # 4 , (4-76)
= p 3 sin # 3 — p 4 sin # 4 , (4-77)
T t = T 3 + Ti + Q. (4-78)
Since kinetic energy can be expressed in terms of momentum, if the masses
are known, we may find any three of the quantities p it p 3 , p 4 , # 3 , # 4 , Q in
terms of the other three. In many cases pi is known, p 3 and «? 3 are meas-
ured, and it is desired to calculate Q. By eliminating # 4 from Eqs. (4-76)
and (4-77), as in the previous example, we obtain
pl = Pi + Pz — 2piPz cos t} 3 . (4-79)
This may now be substituted in Eq. (4-78) to give Q in terms of known
quantities:
f) — T T T — Pl P3 Pl + P3 — 2pip 3 C0S ^3
2mi 2m 3 2m 4 (4_80)
or
« = r. G - 2) - ^ (i + *) + 2 (a^) 1 '^ » 3 .
(4-81)
Every step up to the substitution for T\, T 3 , and r 4 is valid also for par-
ticles moving at velocities of the order of the velocity of light. At high
velocities, the relativistic relation (4-73) between T and p should be used
in the last step. Equation (4-81) is useful in obtaining Q for a nuclear
reaction in which an incident particle mi of known energy collides with a
nucleus m 2 , with the result that a particle m 3 is emitted whose energy and
direction of motion can be observed. Equation (4-81) allows us to deter-
mine Q from these known quantities, taking into account the effect of the
slight recoil of the residual nucleus m 4 , which is usually difficult to observe
directly.
Collisions of inert macroscopic bodies are always inelastic and endoergic,
kinetic energy being converted to heat by frictional forces during the im-
pact. Kinetic energy of translation may also be converted into kinetic
energy of rotation, and conversely. (Exchanges of rotational energy are
included in Q in the previous analysis.) Such collisions range from the
nearly elastic collisions of hard steel balls, to which the above analysis of
elastic collisions applies when rotation is not involved, to completely in-
elastic collisions in which the two bodies stick together after the collision.
178 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
Let us consider a completely inelastic collision in which a bullet of mass
mi, velocity Vi strikes and sticks in an object of mass m 2 at rest. Let the
velocity of the two after the collision be v 2 . Evidently the conservation
of momentum implies that v 2 be in the same direction as Vi, and we have:
miVi = (mi + m 2 )v 2 . (4-82)
The velocity after the collision is
v 2 = ^-Vl (4-83)
mi + m 2
Energy is not conserved in such a collision. The amount of energy con-
verted into heat is
Q = \ mx v\ - i(mx + m 2 )v\ = hm x v\ ( m ™+ m ^ ■ (4-84)
In a head-on collision of two bodies in which rotation is not involved, it
was found experimentally by Isaac Newton that the ratio of relative veloc-
ity after impact to relative velocity before impact is roughly constant for
any two given bodies. Let bodies mi,m 2 , traveling with initial velocities
vn, v 2 i along the z-axis, collide and rebound along the same axis with
velocities Vif, v 2 f. Then the experimental result is expressed by the
equation*
v 2F — vif = e(vir — v 2I ), (4-85)
where the constant e is called the coefficient of restitution, and has a value
between and 1. If e = 1, the collision is perfectly elastic; if e = 0, it is
completely inelastic. Conservation of momentum yields, in any case,
miVu + m 2 v 2 i = miViF + m 2 v 2F . (4-86)
Equations (4-85) and (4-86) enable us to find the final velocities V\ F and
v 2 p for a head-on collision when the initial velocities are known.
4-7 The two-body problem. We consider in this section the motion of a
system of two particles acted on by internal forces satisfying Newton's
third law (weak form), and by no external forces, or by external forces
satisfying a rather specialized condition to be introduced later. We shall
find that this problem can be separated into two single-particle problems.
* More recent experiments show that e is not really constant, but depends on
the initial velocities, on the medium in which the collision takes place, and on
the past history of the bodies. For a more complete discussion with references,
see G. Barnes, "Study of Collisions," Am^ J. Phys. 26, 5 (January, 1958).
4-7] THE TWO-BODY PROBLEM 179
The motion of the center of mass is governed by an equation (4-18) of the
same form as that for a single particle. In addition, we shall find that the
motion of either particle, with respect to the other as origin, is the same
as the motion with respect to a fixed origin, of a single particle of suitably
chosen mass acted on by the same internal force. This result will allow
application of the results of Section 3-14 to cases where the motion of the
attracting center cannot be neglected.
Let the two particles have masses mi and m 2 , and let them be acted on
by external forces Ff , F|, and internal forces F\, F| exerted by each parti-
cle on the other, and satisfying Newton's third law:
Fj = -F 2 . (4-87)
The equations of motion for the system are then
mif j = FJ + F?, (4-88)
m 2 f 2 = F 2 + FJ. (4-89)
We now introduce a change of coordinates:
„ _ rmtt + mar, , (4 _ 90)
(4-91)
r, (4-92)
r, (4-93)
where R is the coordinate of the center of mass, and r is the relative coor-
dinate of m x with respect to m 2 . (See Fig. 4-5.) Adding Eqs. (4-88) and
(4-89) and using Eq. (4-87), we obtain the equation of motion for R:
(mi + m 2 )R = FI + F|. (4-94)
Multiplying Eq. (4-89) by mi, and subtracting from Eq. (4-88) multiplied
xs. -
m
i + m 2
r =
= r x -
- r 2 .
The
inverse
transformation is
ri =
R +
m 2
mi + m 2
r 2 =
R -
mi
mi + m 2
0'
Fig. 4-5. Coordinates for the two-body problem.
180 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
by m 2 , using Eq. (4-87), we obtain the equation of motion for r:
( F e ¥ e \
m^mg = (mi + m 2 )F[ + m x m 2 I — ) • (4-95)
\Wll m 2 /
We now assume that
^- = ^-, (4-96)
■mi m 2
and introduce the abbreviations
M = mi + m 2 , (4-97)
<*= m T 2 ' (4-98)
mi + m 2
F = F\ + F|. (4-99)
Equations (4-94) and (4-95) then take the form of single-particle equa-
tions of motion :
M& = F, (4-100)
fix = F\. (4-101)
Equation (4-100) is the familiar equation for the motion of the center of
mass. Equation (4-101) is the equation of motion for a particle of mass n
acted on by the internal force F\ that particle 2 exerts on particle 1.
Thus the motion of particle 1 as viewed from particle 2 is the same as if
particle 2 were fixed and particle 1 had a mass n (n is called the reduced
mass). If one particle is much heavier than the other, /x is slightly less
than the mass of the lighter particle. If the particles are of equal mass, /*
is half the mass of either. We may now apply the results of Section 3-14
to any two-body problem in which the two particles exert an inverse square
law attraction or repulsion on each other, provided the external forces are
either zero or are proportional to the masses, as required by Eq. (4-96).
Equation (4-96) is satisfied if the external forces are gravitational forces
exerted by masses whose distances from the two bodies mi and m 2 are
much greater than the distance r from mi to m 2 . As an example, the motion
of the earth-moon system can be treated, to a good approximation, by the
method of this section, since the moon is much closer to the earth than
either is to the sun (or to the other planets). Atomic particles are acted
on by electrical forces proportional to their charges, and hence Eq. (4-96)
holds ordinarily only if the external forces are zero. There is also the less
important case where the two particles have the same ratio of charge to
mass, and are acted on by external forces due to distant charges. We may
remark here that although Eqs. (4-88) and (4-89) are not the correct equa-
tions for describing the motions of atomic particles, the introduction of the
4-8] CENTEH-OF-MASS COORDINATES 181
coordinates R, r, and the reduction of the two-body problem to two one-
body problems can be carried out in the quantum-mechanical treatment in
a way exactly analogous to the above classical treatment, under the same
assumptions about the forces.
It is worth remarking that the kinetic energy of the two-body system
can be separated into two parts, one associated with each of the two one-
body problems into which we have separated the two-body problem. The
center-of-mass velocity and the relative velocity are, according to Eqs.
(4-90)-(4-93), related to the particle velocities by
V — R — miVl + m2V2
mi + m 2
(4-102)
v = f = vi — v 2 ,
(4-103)
or
V! = V + -£-V,
mi
(4-104)
v 2 = V - -£- v.
m 2
(4-105)
The total kinetic energy is
T = \m\v\ + |m 2 f I
= iMv 2 + iiw 2 -
(4-106)
The angular momentum can similarly be separated into two parts:
L = mi(ri x v x ) + m 2 (r 2 x v 2 )
= M(R x V) + M(r x v). (4-107)
The total linear momentum is, however, just
P = mxVi + m 2 v 2 = MV. (4-108)
There is no term /*v in the total linear momentum.
4-8 Center-of-mass coordinates. Rutherford scattering by a charged
particle of finite mass. By making use of the results of the preceding sec-
tion, we can solve a two-body scattering problem completely, if we know
the interaction force between the two particles, by solving the one-body
equation of motion for the coordinate r. The result, however, is not in
a very convenient form for application. The solution r(t) describes the
motion of particle 1 with respect to particle 2 as origin. Since particle 2
itself will be moving along some orbit, this is not usually a very convenient
way of interpreting the motion. It would be better to describe the motion
of both particles by means of coordinates r x (0, r 2 (i) referred to some fixed
182 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
origin. Usually one of the particles is initially at rest; we shall take it to
be particle 2, and call it the target particle. Particle 1, approaching the
target with an initial velocity vu, we shall call the incident particle. The
two particles are to be located by vectors r x and r 2 relative to an origin
with respect to which the target particle is initially at rest. We shall call
the coordinates r 1; r 2 the laboratory coordinate system.
The translation from the coordinates R, r to laboratory coordinates is
most conveniently carried out in two steps. We first introduce a center-of-
mass coordinate system in which the particles are located by vectors x\, r 2
with respect to the center of mass as origin:
^ = fl ~ R ' (4-109)
*2 = r 2 — K-i
and, conversely,
ri = r ' 1+R ' (4-110)
r 2 = x\ + R.
The relation between the center-of-mass coordinates and the relative co-
ordinate r is obtained from Eqs. (4-92) and (4-93) :
' n ' + Z ""' , <*-»»
r * mi r = cL-r
2 mi + m 2 m 2 '
The position vectors of the particles relative to the center of mass are con-
stant multiples of the relative coordinate r. The center of mass has the
advantage over particle 2, as an origin of coordinates, in that it moves
with uniform velocity in collision problems where no external forces are
assumed to act.
In the center-of-mass coordinate system the total linear momentum is
zero, and the momenta p\ and p| of the two particles are always equal and
opposite. The scattering angles t?i and d{ between the two final directions
of motion and the initial direction of motion of particle 1 are the supple-
ments of each other, as shown in Fig. 4-6.
We now determine the relation between the scattering angle © in the
equivalent one-body problem and the scattering angle j?i in the laboratory
coordinate system (Fig. 4-7). The velocity of the incident particle in the
center-of-mass system is related to the relative velocity in the one-body
problem, according to Eq. (4-111), by
v{ = -£-v. (4-112)
mi
4-8]
CENTER-OF-MASS COORDINATES
183
Fig. 4-6. Two-particle collision in Fig. 4-7. Orbits for two-body colli-
center-of-mass coordinates. sion in the laboratory system.
Since these two velocities are always parallel, the angle of scattering &[
of the incident particle in the center-of-mass system is equal to the angle
of scattering © in the one-body problem. The incident particle velocities
in the center-of-mass and laboratory systems are related by [Eq. (4-110)]
Vl = vi + V,
(4-113)
where the constant velocity of the center of mass can be expressed in terms
of the initial velocity in the laboratory system by Eq. (4-102) :
V =
mi a
mi + m 2 m 2
(4-114)
The relation expressed by Eq. (4-113) is shown in Fig. 4-8, from which
the relation between d\ = and #i can be determined:
. . v\f sin @
tan#i = — — — ,
v\ F cos S + V
(4-115)
Fig. 4-8. Relation between velocities in laboratory and center-of-mass co-
ordinate systems.
184 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
or, with the help of Eqs. (4-112) and (4-114),
sin /a iia\
tan th = ~ , , t r > (4-116)
cos + (miVi/m 2 VF)
where vz and vp are the initial and final relative speeds, and we have
substituted i>j for vn, since initially the relative velocity is just the velocity
of the incident particle. If the collision is elastic, the initial and final speeds
are the same and Eq. (4-116) reduces to:
sin © /, ,,«
tani?! = „ , , -, — r- (4-117)
1 cos + (wi/m 2 )
A similar relation for t? 2 can be worked out.
If the incident particle is much heavier than the target particle, then
t?i will be very small, no matter what value may have. This corre-
sponds to the result obtained in Section 4-6, that t? x can never be larger
than t> m given by Eq. (4-68), if m l > m 2 . If mi = m 2 , then Eq. (4-117)
is easily solved for #i :
x , sin 2 sin (0/2) cos (0/2) _
tan * 1 = cos + 1 = 2 cos* (0/2) - tan 2 '
0! = J0. (4-118)
Since may always have any value between and w without violating the
conservation laws in the center-of-mass system, the maximum value of «?i
in this case is 7r/2, in agreement with the corresponding result of Section
4-6. If the target mass m 2 is much larger than the incident mass mi, then
tan«?i = tan©; this justifies rigorously our application to this case of
Eq. (3-276) for the Rutherford cross section, deduced in Chapter 3 for
the one-body scattering problem with an inverse square law force.
According to the above developments, Eq. (3-276) applies also to the
two-body problem for any ratio m,i/m 2 of incident mass to target mass,
but must be interpreted as the angle of scattering in terms of relative
coordinates, or else in terms of center-of-mass coordinates. That is, da
in Eq. (3-276) is the cross section for a scattering process in which the
relative velocity v after the collision makes an angle between and © + d©
with the initial velocity. Since it is the laboratory scattering angle #i that
is ordinarily measured, we must substitute for and d© in Eq. (3-276)
their values in terms of #i and d&i as determined from Eq. (4-117). This
is most easily done in case m t = m 2 , when, by Eq. (4-118), the Rutherford
scattering cross section [Eq. (3-276)] becomes
jr = (Ml) 2 15^*1 2t gin di ^ (4 _ n9)
\2fwy sin 4 #i
4-9] THE AT-BODY PROBLEM 185
4-9 The TV-body problem. It would be very satisfactory if we could
arrive at a general method of solving the problem of any number of parti-
cles moving under the forces which they exert on one another, analogous to
the method given in Section 4-7 by which the two-body problem was re-
duced to two separate one-body problems. Unfortunately no such general
method is available for systems of more than two particles. This does not
mean that such problems cannot be solved. The extremely accurate cal-
culations of the motions of the planets represent a solution of a problem
involving the gravitational interactions of a considerable number of bodies.
However, these solutions are not general solutions of the equations of
motion, like the system of orbits we have obtained for the two-body case,
but are numerical solutions obtained by elaborate calculations for specified
initial conditions and holding over certain periods of time. Even the three-
body problem admits of no general reduction, say, to three one-body prob-
lems, or to any other manageable set of equations.
0'
Fig. 4-9. Center-of-mass and internal coordinates of a system of particles.
However, we can partially separate the problem of the motion of a
system of particles into two problems: first, to find the motion of the center
of mass, and second, to find the internal motion of the system, that is, the
motion of its particles relative to the center of mass. Let us define the in-
ternal coordinate vector rjj. of the fcth particle as the vector from the center
of mass to the fcth particle (Fig. 4-9) :
4 = r* - R, k=l,...,N, (4-120)
i k = R + 4, k=l,...,N. (4-121)
In view of the definition (4-14) of the center of mass, the internal co-
ordinates rjj. satisfy the equation
£ m k ri = 0. (4-122)
186 THE MOTION OF A SYSTEM OP PARTICLES [CHAP. 4
We define the center-of-mass velocity and the internal velocities:
V = % (4-123)
vj = it = v fc - V. (4-124)
The total internal momentum of a system of particles (i.e., the momentum
* relative to the center of mass) vanishes by Eq. (4-122) :
*
H :JV
£ m k vt = 0. (4-125)
fc=i
We first show that the total kinetic energy, momentum, and angular
momentum can each be split up into a part depending on the total mass M
and the motion of the center of mass, and an internal part depending only
on the internal coordinates and velocities. The total kinetic energy of the
system of particles is
T=22 im*»f. (4-126)
&=i
By substituting for v& from Eq. (4-124), and making use of Eq. (4-125),
we can split T into two parts:
JV
T =22 W^ 2 + 2V-v£ + vi a )
k=i
= 22 2 m * v2 + E w* 8 + 12 w * v>v *
&=i fc=i £=1
= \MV 2 + J2 \m k vf + v-22 »*▼*
k=l fc=l
= \MV 2 + 22 i m ^l 2 - (4-127)
The total linear momentum is, if we make use of Eqs. (4-124) and
p = 22 mkVk
iV N
= 22 m * v + 2 mkV ' c
k=l fc=l
= MV. (4-128)
The internal linear momentum is zero.
4-9] THE 2V-BODY PROBLEM 187
The total angular momentum about the origin is, if we use Eqs. (4-121),
(4-122), (4-124), and (4-125),
L = ^2 m k (r k X Vfc)
N
= J2 m *( R x V + 4 x V + R x v| + 4 x vl)
= J2 m *( R xV) + (t ™* r * J x V + R x ( X) m * v * J
fc=l \fc=l / \fc=l /
I
fc=l
+ 53 m M x ▼*)
N
I
= M(R X V) + £ »*('* x v *)- (4^129)
Notice that the internal angular momentum depends only on the internal
coordinates and velocities and is independent of the origin about which L
is being computed (and from which the vector R is drawn).
The position of particle k with respect to particle I is specified by the
vector
r* - r, = 4 - rf. (4-130)
The relative positions of the particles with respect to each other depend
only on the internal coordinates r^, and likewise the relative velocities, so
that the internal forces Ff. will be expected to depend only on the internal
coordinates x\, and possibly on the internal velocities. If there is a poten-
tial energy associated with the internal forces, it likewise will depend only
on the internal coordinates.
Although the forces, energy, momentum, and angular momentum can
each be split into two parts, a part associated with the motion of the center
of mass and an internal part depending only on the internal coordinates and
velocities, it must not be supposed that the internal motion and the center-
of-mass motion are two completely separate problems. The motion of the
center of mass, as governed by Eq. (4-18), is a separate one-body problem
when the external force F is given. However, in most cases F will depend
to some extent on the internal motion of the system. The internal equa-
tions of motion contain the external forces except in special cases and,
furthermore, they also depend on the motion of the center of mass. If we
substitute Eqs. (4-121) in Eqs. (4-1), and rearrange, we have
m k H = Pi + F£ - m fc R. (4-131)
188 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
There are many cases, however, in which a group of particles forms a
system which seems to have some identity of its own independent of other
particles and systems of particles. An atomic nucleus, made up of neutrons
and protons, is an example, as is an atom, made up of nucleus and electrons,
or a molecule, composed of nuclei and electrons, or the collection of particles
which make up a baseball. In all such cases, it turns out that the internal
forces are much stronger than the external ones, and the acceleration R is
small, so that the internal equations of motion (4-131) depend essentially
only on the internal forces, and their solutions represent internal motions
which are nearly independent of the external forces and of the motion of
the system as a whole. The system viewed externally then behaves like
a single particle with coordinate vector R, mass M, acted on by the
(external) force F, but a particle which has, in addition to its "orbital"
energy, momentum, and angular momentum associated with the motion
of its center of mass, an intrinsic or internal energy and angular momentum
associated with its internal motion. The orbital and intrinsic parts of the
energy, momentum, and angular momentum can be identified in Eqs.
(4-127), (4-128), and (4-129). The internal angular momentum is usually
called spin and is independent of the position or velocity of the center of
mass relative to the origin about which the total angular momentum is to
be computed. So long as the external forces are small, this approximate
representation of the system as a single particle is valid. Whenever the
external forces are strong enough to affect appreciably the internal motion,
the separation into problems of internal and of orbital motions breaks
down and the system begins to lose its individuality. Some of the central
problems at the frontiers of present-day physical theories are concerned
with bridging the gap between a loose collection of particles and a system
with sufficient individuality to be treated as a single particle.
4-10 Two coupled harmonic oscillators. A very commonly occurring
type of mechanical system is one in which several harmonic oscillators
interact with one another. As a typical example of such a system, con-
sider the mechanical system shown in Fig. 4-10, consisting of two masses
mi, m 2 fastened to fixed supports by springs whose elastic constants are
fci, k 2 , and connected by a third spring of elastic constant £3. We suppose
the masses are free to move only along the rc-axis; they may, for example,
slide along a rail. If spring A; 3 were not present, the two masses would
1
U-Zl-^l |~ X 2-]
fa ] mi jfc 3 mi ; k 2
Fig. 4-10. A simple model of two coupled harmonic oscillators.
4-10] TWO COUPLED HARMONIC OSCILLATORS 189
vibrate independently in simple harmonic motion with angular frequencies
(neglecting damping)
.0 [k~i /&2
* = Vv w - = v^- (4 " 132)
We wish to investigate the effect of coupling these two oscillators to-
gether by means of the spring k 3 . We describe the positions of the two
masses by specifying the distances xi and x 2 that the springs ki and k 2
have been stretched from their equilibrium positions. We assume for
simplicity that when springs fci and k 2 are relaxed (x\ = x 2 = 0), spring
k 3 is also relaxed. The amount by which spring fc 3 is compressed is then
(xi + x 2 )- The equations of motion for the masses mi, m 2 (neglecting
friction) are
miXi = — kiXi — k 3 (xi + x 2 ), (4-133)
m 2 x 2 = —k 2 x 2 — k 3 (xi + x 2 ).
(4-134)
he form
miXi + k'lXi + k a x 2 = 0,
(4-135)
m 2 x 2 + ¥ 2 x 2 + k 3 x x = 0,
(4-136)
k\ = *i + ^3,
(4-137)
k' 2 = k 2 + ^3-
(4-138)
where
We have two second-order linear differential equations to solve simul-
taneously. If the third terms were not present, the equations would be
independent of one another, and we would have independent harmonic
vibrations of X\ and x 2 at frequencies
«io = >/J> (4-139)
fcT
W20 = v^ • ^ 14 °)
These are the frequencies with which each mass would vibrate if the other
were held fixed. Thus the first effect of the coupling spring is simply to
change the frequency of independent vibration of each mass, due to the
fact that each mass is now held in position by two springs instead of one.
The third terms in Eqs. (4-135) and (4-136) give rise to a coupling between
the motions of the two masses, so that they no longer move independently.
190 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
We may solve Eqs. (4-135), (4-136) by an extension of the method of
Section 2-8 applicable to any set of simultaneous linear differential equa-
tions with constant coefficients. We assume that
Xx = C l e pt , t (4-141)
x 2 = C 2 e pt , (4-142)
where Ci, C 2 are constants. Note that the same time dependence is as-
sumed for both Xi and x 2 , in order that the factor e pt will cancel out when
we substitute in Eqs. (4-135) and (4-136) :
(m lP 2 + kUd + k 3 C 2 = 0, (4-143)
(m 2 p 2 + k' 2 )C 2 + k 3 d = 0. (4-144)
We now have two algebraic equations in the three unknown quantities
C\, C 2 , p. We note that either Eq. (4-143) or (4-144) can be solved for
the ratio C 2 /C\ :
Cz = _ m x p 2 + fc'i = _ k 3 _ (4-145)
Ci k 3 m 2 p 2 + k 2
The two values of C 2 /Ci must be equal, and we have an equation for p:
m *P 2 + *1 = *2 , (4-146)
k 3 m 2 p 2 + fc'2
which may be rearranged as a quadratic equation in p 2 , called the secular
equation:
m x m 2 p* + (m 2 k\ + mi k' 2 )p 2 + (k\k' 2 - fc|) = 0, (4-147)
whose solutions are
«* = _ if *L + *£} ± [if*L + *kY _ -MMl , fc| 1 1/2
2 \mi m 2 J l4 \mi m 2 ) mim 2 m\m 2 \
I (,,%. _i_ ,..?^ j. Ti /,..?_ _ ,.<?.\ 2 _i_ fc 3 ] 1/2
m\m 2 \
[i<«s.
(«f + «ao) ± 7 («io - oioV + zPtr ■ (4-148)
It is not hard to show that the quantity in brackets is less than the square
of the first term, so that we have two negative solutions for p 2 . If we
assume that «i > w 2 o, the solutions for p 2 are
p 2 = _ w 2 = _ (w?o + i Aa)2)j
2 2 • , 2 1 A Sv (4-149)
P = — « 2 = — («20 — 2 A&> ),
4-10]
where
Aw 2 = (w?o — w|o)
with the abbreviation
TWO COUPLED HARMONIC OSCILLATORS
\ («fo - «Io)V
fca
191
(4-150)
(4-151)
\/mim2
where k is the coupling constant. If «i = «2o> Eq. (4-150) reduces to
Aco 2 = 2k 2 . (4-152)
The four solutions for p are
p = ±r'coi, ±K02- (4-153)
If p 2 = — cof, Eq. (4-145) can be written
C 2 mi , 2 2 n Aw
— (W], — WioJ =
Ci
&s
Mi
2/c 2 "\ m2 :
and if p 2 = — cof, it can be written
m 2
Ci m 2 , 2 2 \ Aa>
(4-154)
(4-155)
By substituting from Eq. (4-153) in Eqs. (4-141), (4-142), we get four
solutions of Eqs. (4-135) and (4-136) provided the ratio C 2 /Ci is chosen
according to Eq. (4-154) or (4-155). Each of these solutions involves one
arbitrary constant (Ci or C 2 ). Since the equations (4-135), (4-136) are
linear, the sum of these four solutions will also be a solution, and is in fact
the general solution, for it will contain four arbitrary constants (say
^D ^D ^2> ^2/ '
1 2k 2 \toi 2/c 2 \mi
(4-156)
Aco* /mi n j ai t , Aco l m Ar</„—i<'it_i_r<.^ i '"2tj_r<'.o— iu '2t
(4-157)
(4-158)
(4-159)
* 2= 2^>fe C ' ie
+ ^ yj^- C"ie- tUlt + C 2 e lM2t + C' 2 e-
In order to make xi and £2 real, we choose
C 2 = iA 2 e J ' 92 , C" 2 = i^e - * 2 ,
192 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
so that
A 2 I
x x = Ai cos («i< + 6{) — 2^- -J^ 2 - A 2 cos (o>2< + B 2 ), (4-160)
x 2 = ^- J2± Ai cos fat + 0i) + A 2 cos (w 2 < + 2 )- (4-161)
This is the general solution, involving the four arbitrary constants
Ai, A 2 , di, 6 2 . We see that the motion of each coordinate is a super-
position of two harmonic vibrations at frequencies «i and o> 2 . The os-
cillation frequencies are the same for both coordinates, but the relative
amplitudes are different, and are given by Eqs. (4-154) and (4-155).
If Ax or A 2 is zero, only one frequency of oscillation appears. The re-
sulting motion is called a normal mode of vibration. The normal mode of
highest frequency is given by
x t = Ai cos (wi* + 0i), (4-162)
* a = i£ VS A * cos (wi< + 6l) ' (4_163)
d = wfo + iAw 2 . (4-164)
The frequency of oscillation is higher than «i . By referring to Fig. 4-10,
we see that in this mode of oscillation the two masses mi and m 2 are oscil-
lating out of phase; that is, their displacements are in opposite directions.
The mode of oscillation of lower frequency is given by
Xi = ~ i£ -n/S a * cos (W2< + 02) ' (4_165)
x 2 = A 2 cos (u 2 t + 9 2 ), (4-166)
o>! = wlo — iAw 2 . (4-167)
In this mode, the two masses oscillate in phase at a frequency lower
than w 2 o- The most general motion of the system is given by Eqs. (4-160),
(4-161), and is a superposition of the two normal modes of vibration.
The effect of coupling is thus to cause both masses to participate in the
oscillation at each frequency, and to raise the highest frequency and lower
the lowest frequency of oscillation. Even when both frequencies are
initially equal, the coupling results in two frequencies of vibration, one
higher and one lower than the frequency without coupling. When the
coupling is very weak, i.e., when
k 2 « Kcofo - «io), (4-168)
4-10] TWO COUPLED HARMONIC OSCILLATORS 193
then Eq. (4-150) becomes
Ac ° 2 - 2 2K 2 • (4- 169 )
w 10 — W20
For the highest frequency mode of vibration, the ratio of the amplitude of
vibration of mass m 2 to that of mass mi is then
x 2 Aco 2 lm[ . k 2
Thus, unless w 2 « Bi, the mass m 2 oscillates at much smaller amplitude
than TO X . Similarly, it can be shown that for the low-frequency mode of
vibration, m x oscillates at much smaller amplitude than m 2 . If two oscil-
lators of different frequency are weakly coupled together, there are two
normal modes of vibration of the system. In one mode, the oscillator of
higher frequency oscillates at a frequency slightly higher than without
coupling, and the other oscillates weakly out of phase at the same fre-
quency. In the other mode, the oscillator of lowest frequency oscillates at
a frequency slightly lower than without coupling, and the other oscillates
weakly and in phase at the same frequency. At or near resonance, when
the two natural frequencies coi and u 20 are equal, the condition for weak
coupling [Eq. (4-168)] is not satisfied even when the coupling constant is
very small. Aco 2 is then given by Eq. (4-152), and we find for the two nor-
mal modes of vibration :
(4-171)
w 2 = w? ± k 2 . (4-172)
The two oscillators oscillate in or out of phase with an amplitude ratio
depending only on their mass ratio, and with a frequency higher or lower
than the uncoupled frequency by an amount depending on the coupling
constant.
An interesting special case is the case of two identical oscillators
(mi = m 2 , hi = k 2 ) coupled together. The general solution (4-160),
(4-161) is, in this case,
Xi — Ai cos (wi< + 81) — A 2 cos (co 2 < + 2 ), (4-173)
x 2 = Ai cos (wi< + 0i) + A 2 cos (« 2 * + #2)) (4-174)
where «i and « 2 are given by Eq. (4-172). If A 2 = 0, we have the high-
frequency normal mode of vibration, and if Ai = 0, we have the low-fre-
quency normal mode. Let us suppose that initially m 2 is at rest in its
equilibrium position, while mi is displaced a distance A from equilibrium
194 THE MOTION OF A SYSTEM OP PARTICLES [CHAP. 4
and released at t = 0. The choice of constants which fits these initial
conditions is
61 = 2 = 0,
Ai = —A 2 = %A,
(4-175)
so that Eqs. (4-173), (4-174) become
xi = JA (coswi* + COSW2O, (4-176)
x 2 = \A (coswi* — cosw 2 <), (4-177)
which can be rewritten in the form
Xl = A cos ("Up* t ) cos (^Jp> f ) , (4-178)
x 2 = -A sin (e^p> ,) sin («L±«S ,) . (4-179)
If the coupling is small, wi and w 2 are nearly equal, and xi and a; 2 oscillate
rapidly at the angular frequency («i + w 2 )/2 == wi = w 2 , with an am-
plitude which varies sinusoidally at angular frequency (w x — w 2 )/2. The
motion of each oscillator is a superposition of its two normal-mode motions,
which leads to beats, the beat frequency being the difference between the
two normal-mode frequencies. This is illustrated in Fig. 4-11, where os-
cillograms of the motion of x 2 are shown: (a) when the high-frequency
normal mode alone is excited, (b) when the low-frequency normal mode
is excited, and (c) when oscillator m x alone is initially displaced. In Fig.
4-12, oscillograms of Xi and x 2 as given by Eqs. (4-178), (4-179) are
shown. It can be seen that the oscillators periodically exchange their
energy, due to the coupling between them. Figure 4-13 shows the same
motion when the springs fci and k 2 are not exactly equal. In this case,
oscillator mi does not give up all its energy to ra 2 during the beats. Figure
4-14 shows that the effect of increasing the coupling is to increase the beat
frequency «i — co 2 [Eq. (4-172)].
If a frictional force acts on each oscillator, the equations of motion
(4-135) and (4-136) become
miXi + bi±! + kiXt + k 3 x 2 = 0, (4-180)
m 2 x 2 + b 2 x 2 + k' 2 x 2 + k 3 xi — 0, (4-181)
4-10]
TWO COUPLED HARMONIC OSCILLATORS
195
(a) lulfiikjtii
Fig. 4-11. Motion of coupled har- Fig. 4-12. Motion of two identical
monic oscillators, (a) High-frequency coupled oscillators,
normal mode, (b) Low-frequency nor-
mal mode, (c) mi initially displaced.
Fig. 4-13. Motion of two nonidenti- Fig. 4-14. Motion of two coupled
cal coupled oscillators. oscillators, (a) Weak coupling, (b)
Strong coupling.
196 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
where b% and b 2 are the respective friction coefficients. The substitution
(4-141), (4-142) leads to a fourth-degree secular equation for p:
m 1 m 2 p 4: + (m 2 bi + mib 2 )p 3 + (m 2 fci + mik' 2 + b x b 2 )p 2
+ (&i*i + b 2 k\)p + (k\k' 2 - fc|) = 0. (4-182)
This equation cannot be solved so easily as Eq. (4-147). The four roots
for p are, in general, complex, and have the form (if &i and b 2 are not too
large)
p = — 7i ± t'wi,
(4-183)
p = —y 2 ± iu 2 .
That the roots have this form with Ti and T 2 positive can be shown (though
not easily) algebraically from a study of the coefficients in Eq. (4-182).
Physically, it is evident that the roots have the form (4-183), since this will
lead to damped vibrations, the expected result of friction. If 6i and b 2 are
large enough, one or both of the pairs of complex roots may become a pair
of real negative roots, the corresponding normal mode or modes being
overdamped. A practical solution of Eq. (4-182) can, in general, be ob-
tained only by numerical methods when numerical values for the constants
are given, although an approximate algebraic solution can be found when
the damping is very small.
The problem of the motion of a system of two coupled harmonic oscil-
lators subject to a harmonically oscillating force applied to either mass can
be solved by methods similar to those which apply to a single harmonic
oscillator. A steady-state solution can be found in which both oscillators
oscillate at the frequency of the applied force with definite amplitudes and
phases, depending on their masses, the spring constants, the damping, and
the amplitude and phase of the applied force. The system is in resonance
with the applied force when its frequency corresponds to either of the two
normal modes of vibration, and the masses then vibrate at large amplitudes
limited only by the damping. The general solution consists of the steady-
state solution plus the general solution of the unforced problem. A super-
position principle can be proved according to which, if a number of forces
act on either or both masses, the solution is the sum of the solutions with
each force acting separately. This theorem can be used to treat the prob-
lem of arbitrary forces acting on the two masses.
Other types of coupling between the oscillators are possible in addition
to coupling by means of a spring as in the example above. The oscillators
may be coupled by frictional forces. A simple example would be the case
where one mass slides over the other, as in Fig. 4-15. We assume that the
force of friction is proportional to the relative velocity of the two masses.
4-10]
TWO COUPLED HARMONIC OSCILLATORS
197
u_n— J
KM
mi
mo i i ■"
-^2 — ^ *'2
Fig. 4-15. Frictional coupling.
-h— "->
t3;
-x 2
-A
Fig. 4-16. Coupling through a mass.
The equations of motion of mi and m 2 are then
m^i = — k\Xi — b{±i + x 2 ),
m 2 x 2 = —k 2 x 2 — b(x 2 + ±i),
or
miXi + b±x + kiXi + bx 2 = 0,
w 2 x 2 + bx 2 + k 2 x 2 + &±! = 0.
(4-184)
(4-185)
(4-186)
(4-187)
The coupling is expressed in Eqs. (4-186), (4-187) by a term in the equation
of motion of each oscillator depending on the velocity of the other. The
oscillators may also be coupled by a mass, as in Fig. 4-16. It is left to the
reader to set up the equations of motion. (See Problem 26 at the end of
this chapter.)
Two oscillators may be coupled in such a way that the force acting on
one depends on the position, velocity, or acceleration of the other, or on
any combination of these. In general, all three types of coupling occur to
some extent; a spring, for example, has always some mass, and is subject
to some internal friction. Thus the most general pair of equations for two
coupled harmonic oscillators is of the form
m x x\ + bi±i + k x xi + m c x 2 + b c x 2 + k c x 2 = 0, (4-188)
m 2 x 2 + b 2 x 2 + k 2 x 2 + m c xi + b e x\ + k c x\ = 0. (4-189)
These equations can be solved by the method described above, with similar
results. Two normal modes of vibration appear, if the frictional forces
are not too great.
Equations of the form (4-188), (4-189), or the simpler special cases con-
sidered in the preceding discussions, arise not only in the theory of coupled
mechanical oscillators, but also in the theory of coupled electrical circuits.
Applying Kirchhoff 's second law to the two meshes of the circuit shown in
Fig. 4-17, with mesh currents i\, i 2 around the two meshes as shown, we
obtain
(L + Li)g! + (R + Rtfa +
fe + £) qi
1
+ Lq 2 + Rq 2 + ^ q 2 = 0,
(4-190)
198
THE MOTION OF A SYSTEM OF PARTICLES
Ki R 2
[chap. 4
Ci C 2
Fig. 4-17. Coupled oscillating circuits.
and
(L + L 2 )q 2 + (R+ R 2 )q 2 +
(h+£
q 2 + Lqt + Rqi+jjqi = 0,
(4-191)
where qi and q 2 are the charges built up on Ci and C 2 by the mesh currents
ii and i 2 . These equations have the same form as Eqs. (4-188), (4-189),
and can be solved by similar methods. In electrical circuits, the damping
is often fairly large, and finding the solution becomes a formidable task.
The discussion of this section can be extended to the case of any number
of coupled mechanical or electrical harmonic oscillators, with analogous
results. The algebraic details become almost prohibitive, however, unless
we make use of more advanced mathematical techniques. We therefore
postpone further discussion of this problem to Chapter 12.
All mechanical and electrical vibration problems reduce in the limiting
case of small amplitudes of vibration to problems involving one or several
coupled harmonic oscillators. Problems involving vibrations of strings,
membranes, elastic solids, and electrical and acoustical vibrations in trans-
mission lines, pipes, or cavities, can be reduced to problems of coupled
oscillators, and exhibit similar normal modes of vibration. The treatment
of the behavior of an atom or molecule according to quantum mechanics
results in a mathematical problem identical with the problem of coupled
harmonic oscillators, in which the energy levels play the role of oscillators,
and external perturbing influences play the role of the coupling mechanism.
199
Problems
1. Formulate and prove a conservation law for the angular momentum about
the origin of a system of particles confined to a plane.
2. Water is poured into a barrel at the rate of 120 lb per minute from a height
of 16 ft. The barrel weighs 25 lb, and rests on a scale. Find the scale reading
after the water has been pouring into the barrel for one minute.
3. A scoop of mass mi is attached to an arm of length I and negligible weight.
The arm is pivoted so that the scoop is free to swing in a vertical arc of radius I.
At a distance I directly below the pivot is a pile of sand. The scoop is lifted until
the arm is at a 45° angle with the vertical, and released. It swings down and
scoops up a mass m,2 of sand. To what angle with the vertical does the arm of the
scoop rise after picking up the sand? This problem is to be solved by considering
carefully which conservation laws are applicable to each part of the swing of the
scoop. Friction is to be neglected, except that required to keep the sand in the
scoop.
4. (a) A spherical satellite of mass m, radius a, moves with speed v through
a tenuous atmosphere of density p. Find the frictional force on it, assuming
that the speed of the air molecules can be neglected in comparison with v,
and that each molecule which is struck becomes embedded in the skin of the
satellite, (b) If the orbit is a circle 400 km above the earth (radius 6360 km),
where p = 10 -11 kgm/m -3 , and if a = 1 m, m = 100 kgm, find the change
in altitude and the change in period of revolution in one week.
5. A two-stage rocket is to be built capable of accelerating a 100-kgm payload
to a velocity of 6000 m/sec in free flight. (In a two-stage rocket, the first stage
is detached after exhausting its fuel, before the second stage is fired.) Assume
that the fuel used can reach an exhaust velocity of 1500 m/sec, and that struc-
tural requirements imply that an empty rocket (without fuel or payload) will
weigh 10 % as much as the fuel it can carry. Find the optimum choice of masses
for the two stages so that the total take-off weight is a minimum. Show that it
is impossible to build a single-stage rocket which will do the job.
6. A rocket is to be fired vertically upward. The initial mass is Mo, the exhaust
velocity — u is constant, and the rate of exhaust — (dM/dt) = A is constant.
After a total mass A M is exhausted, the rocket engine runs out of fuel. Neglect-
ing air resistance and assuming that the acceleration g of gravity is constant,
set up and solve the equation of motion, and show that if Mo, u, and AM are
fixed, then the larger the rate of exhaust A, that is, the faster it uses up its fuel,
the greater the maximum altitude reached by the rocket.
7. A uniform spherical planet of radius a revolves about the sun in a circular
orbit of radius ro, and rotates about its axis with angular velocity coo, normal
to the plane of the orbit. Due to tides raised on the planet, its angular velocity
of rotation is decreasing. Find a formula expressing the orbit radius r as a func-
tion of angular velocity to of rotation at any later or earlier time. [You will
need formulas (5-9) and (5-91) from Chapter 5.] Apply your formula to the
earth, neglecting the effect of the moon, and estimate how much farther the
earth will be from the sun when the day has become equal to the present year.
If the effect of the moon were taken into account, would the distance be greater
or less?
200 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
*8. A mass m of gas and debris surrounds a star of mass M . The radius of the
star is negligible in comparison with the distances to the particles of gas and
debris. The material surrounding the star has initially a total angular momentum
L, and a total kinetic and potential energy E. Assume that m « Jlf , so that
the gravitational fields due to the mass m are negligible in comparison with that
of the star. Due to internal friction, the surrounding material continually loses
mechanical energy. Show that there is a maximum energy AE which can be lost
in this way, and that when this energy has been lost, the material must all lie
on a circular ring around the star (but not necessarily uniformly distributed).
Find AE and the radius of the ring. (You will need to use the method of La-
grange multipliers.)
9. A particle of mass mi, energy Tu collides elastically with a particle of mass
m,2, at rest. If the mass «2 leaves the collision at an angle #2 with the original
direction of motion of mi, find the energy T%f delivered to particle m2. Show
that T2F is a maximum for a head-on collision, and that in this case the energy
lost by the incident particle in the collision is
T T 4miw 2 „
Tu ~ TlF = (mi + m 2 )2 Tu -
10. A cloud-chamber picture shows the track of an incident particle which
makes a collision and is scattered through an angle #1. The track of the target
particle makes an angle #2 with the direction of the incident particle. Assuming
that the collision was elastic and that the target particle was initially at rest, find
the ratio mi/wi2 of the two masses. (Assume small velocities so that the classical
expressions for energy and momentum may be used.)
11. Show that an elastic collision corresponds to a coefficient of restitution
e = 1, that is, show that for a head-on elastic collision between two particles,
Eq. (4-85) holds with e = 1.
12. Calculate the energy loss Q for a head-on collision between a particle of
mass mi, velocity f 1 with a particle of mass wi2 at rest, if the coefficient of restitu-
tion is e.
13. A particle of mass mi, momentum pu collides elastically with a particle
of mass m2, momentum P21 going in the opposite direction. If mi leaves the colli-
sion at an angle <&i with its original course, find its final momentum.
14. Find the relativistic corrections to Eq. (4-81) when the incident particle
mi and the emitted particle W13 move with speeds near the speed of light. Assume
that the recoil particle mi is moving slowly enough so that the classical relation
between energy and momentum can be used for it.
15. A particle of mass mi, momentum pi collides with a particle of mass m,2 at
rest. A reaction occurs from which two particles of masses WI3 and mi result,
which leave the collision at angles #3 and #4 with the original path of mi. Find
the energy Q absorbed in the reaction in terms of the masses, the angles, and pi.
16. A nuclear reaction whose Q is known occurs in a photographic plate in
which the tracks of the incident particle mi and the two product particles m3
and m4 can be seen. Find the energy of the incident particle in terms of mi,
m3, m4, Q, and the measured angles #3 and #4 between the incident track and
the two final tracks. What happens if Q = 0?
PROBLEMS 201
17. The Compton scattering of x-rays can be interpreted as the result of elastic
collisions between x-ray photons and free electrons. According to quantum
theory, a photon of wavelength X has a kinetic energy hc/\, and a linear momen-
tum of magnitude h/\, where h is Planck's constant and c is the speed of light.
In the Compton effect, an incident beam of x-rays of known wavelength Xj in a
known direction is scattered in passing through matter, and the scattered radia-
tion at an angle #1 to the incident beam is found to have a longer wavelength \f,
which is a function of the angle t?i- Assuming an elastic collision between an inci-
dent photon and an electron of mass m at rest, set up the equations expressing
conservation of energy and momentum. Use the relativistic expressions for the
energy and momentum of the electron. Show that the change in x-ray wave-
length is
Xf — X/ = — (1 — cos #i),
mc
and that the ejected electron appears at an angle given by
sin j?i
tan #2 =
[1+ (h/\imc)](l — cos^x)
18. Work out a correction to Eq. (3-267) which takes into account the motion
of the central mass M under the influence of the revolving mass m. A pair of stars
revolve about each other, so close together that they appear in the telescope as a
single star. It is determined from spectroscopic observations that the two stars
are of equal mass and that each revolves in a circle with speed v and period r
under the gravitational attraction of the other. Find the mass m of each star by
using your formula.
19. Show that if the incident particle is much heavier than the target particle
(wi 2> m,2), the Rutherford scattering cross section da [Eq. (3-276)] in laboratory
coordinates is approximately
dc ± I _£i^ J 2 i /22 2 o 1,2 2ir sin #1 d #i
\2m 2 vV [1 - (1 - 7 2 # 2 ) 1/2 ] 2 (1 - 7 2 # 2 ) 1/2
if 7#i < 1, where 7 = mi/m.2- Otherwise, da = 0.
20. Find an expression analogous to Eq. (4-116) for the angle of recoil of the
target particle (#2 in Fig. 4-7) in terms of the scattering angle in the equivalent
one-body problem. Show that, for an elastic collision,
$2 = 4(t - 0).
21. Assume that m,2 2> mi, and that = #1 + 5, in Eq. (4-117). Find a
formula for 5 in terms of t?i. Show that the first-order correction to the Ruther-
ford scattering cross section [Eq. (3-276)], due to the finite mass of m,2, vanishes.
22. Set up the equations of motion for Fig. 4-10, assuming that the relaxed
length of each spring is I, and that the distance between the walls is 3(Z + a), so
that the springs are stretched, even in the equilibrium position. Show that the
equations can be put in the same form as Eqs. (4-135) and (4-136).
202 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4
23. For the normal mode of vibration given by Eqs. (4-162) and (4-163), find
the force exerted on mi through the coupling spring, and show that the motion
of xi satisfies the equation for a simple harmonic oscillator subject to this driving
force.
24. The system of coupled oscillators shown in Fig. 4-10 is subject to an ap-
plied force
F = Fo cos cot,
applied to mass mi. Set up the equations of motion and find the steady-state
solution. Sketch the amplitude and phase of the oscillations of each oscillator as
functions of w.
25. Find the two normal modes of vibration for a pair of identical damped
coupled harmonic oscillators [Eqs. (4-180), (4-181)]. That is, mi = m2, &i = 62,
fci = fe. [Hint: If fe = 0, you can certainly find the solution. You will find this
point helpful in factoring the secular equation.]
26. Set up the equations of motion for the system shown in Fig. 4-16. The
relaxed lengths of the two springs are h, h- Separate the problem into two
problems, one involving the motion of the center of mass, and the other involving
the "internal motion" described by the two coordinates x\, X2- Find the normal
modes of vibration.
CHAPTER 5
RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
5-1 The dynamical problem of the motion of a rigid body. In order to
apply the theorems of the preceding chapter to the motion of a rigid body,
we regard a rigid body as a system of many particles whose positions rela-
tive to one another remain fixed. We may define a rigid body as a system
of particles whose mutual distances are all constant. The forces which
hold the particles at fixed distances from one another are internal forces,
and may be imagined as exerted by rigid weightless rods connected be-
tween all pairs of particles. Forces like this which maintain certain fixed
relations between the particles of a system are called forces of constraint.
Such forces of constraint can always be regarded as satisfying Newton's
third law (strong form), since the constraints could be maintained by rigid
rods fastened to the particles by frictionless universal joints. We may
therefore apply the theorems of conservation of linear and angular momen-
tum to the motion of a rigid body. For a perfectly rigid body, the theorem
of conservation of mechanical energy holds also, since we can show by
Newton's third law that the forces of constraint do no work in a rigid mo-
tion of the system of particles. The work done by the force exerted by a
moving rod on a particle at one end is equal and opposite to the work done
by the force exerted by the rod on a particle at the other end, since both
particles have the same component of velocity in the direction of the rod
(Fig. 5-1):
F 2 -»i-vi + Fi_ 2 -v 2 = F 2 _rvi — F 2 _»i-v 2 (5-1)
= F 2 _ > i-(vi— v 2 )
= 0.
We shall base our derivation of the equations of motion of a rigid body
on these conservation laws. No actual solid body is ever perfectly rigid,
so that our theory of the motion of rigid bodies will be an idealized ap-
proximation to the motion of actual bodies. However, in most applica-
tions the deviation of actual solid bodies from true rigidity is not sig-
nificant. In a like spirit is our assumption that the ideal rigid body can
be imagined as made up of ideal point particles held at fixed distances
from one another.
A solid body of ordinary size is composed of such a large number of
atoms and molecules that for most purposes it is more convenient to repre-
203
204 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
Fig. 5-1. Forces exerted by two particles connected by a rigid rod.
sent its structure by specifying the average density p of mass per unit
volume at each point in the body. The density is defined by
P =
dM
dV' 1
(5-2)
where dM is the total mass in a volume dV which is to be chosen large
enough to contain a large number of atoms, yet small enough so that the
properties of the material are practically uniform within the volume dV.
Only when a, dV satisfying these two requirements can be chosen in the
neighborhood of a point in the body can the density p be properly defined
at that point. Sums over all the particles, such as occur in the expressions
for total mass, total momentum, etc., can be replaced by integrals over the
volume of the body. For example, the total mass is
M^J^rm^ fff p
dV.
(5-3)
(body)
Further examples will appear in the following sections.
In order to describe the position of a rigid body in space, six coordinates
are needed. We may, for example, specify the coordinates (x lt y lt z{) of
some point Pi in the body. Any other point P 2 of the body a distance r
from Pi will then lie somewhere on a sphere of radius r with center at
(£11 2/1, zi). We can locate P 2 on this sphere with two coordinates, for
example, the spherical coordinate angles 2 , <p 2 with respect to a set of
axes through the point tei, j/i, z\). Any third point P 3 a distance a^O
from the line through Pi and P 2 must now lie on a circle of radius a about
this line. We can locate P3 on this circle with one coordinate. We thus
require a total of six coordinates to locate the three points Pi, P 2 , P3 of
5-1] THE DYNAMICAL PROBLEM OF THE MOTION OF A RIGID BODY 205
the body, and when three noncollinear points are fixed, the locations of
all points of a rigid body are fixed. There are many possible ways of choos-
ing six coordinates by which the position of a body in space can be specified.
Usually three of the six coordinates are used as above to locate some point
in the body. The remaining three coordinates determine the orientation
of the body about this point.
If a body is not connected to any supports, so that it is free to move in
any manner, it is convenient to choose the center of mass as the point to
be located by three coordinates (X, Y, Z), or by the vector R. The motion
of the center of mass R is then determined by the linear momentum theo-
rem, which can be expressed in the form (4-18) :
MR = F, (5-4)
where M is the total mass and F is the total external force. The equation
for the rotational motion about the center of mass is given by the angular
momentum theorem (4-28) :
Tt = N ' <™>
where L is the angular momentum and N is the torque about the point R.
If the force F is independent of the orientation of the body in space, as in
the case of a body moving in a uniform gravitational field, the motion of
the center of mass is independent of the rotational motion, and Eq. (5-4)
is a separate equation which can be solved by the methods of Chapter 3.
If the torque N is independent of the position R of the center of mass, or
if R(t) is already known, so that N can be calculated as a function of time
and of the orientation of the body, then the rotational motion about the
center of mass may be determined from Eq. (5-5). In the more general
case, when F and N each depend on both position and orientation, Eqs.
(5-4) and (5-5) must be solved simultaneously as six coupled equations
in some suitable set of coordinates; this case we shall not attempt to treat,
although after the reader has studied Chapter 11, he will be able to set up
for himself the six equations which must be solved.
If the body is constrained by external supports to rotate about a fixed
point 0, then moments and torques are to be computed about that point.
We have to solve Eq. (5-5) for the rotation about the point 0. In this case
Eq. (5-4) serves only to determine the constraining force required to
maintain the point at rest.
The difficulty in applying Eq. (5-5) lies in the choice of three coordinates
to describe the orientation of the body in space. The first thought that
comes to mind is to choose a zero position for the body, and to specify
any other orientation by specifying the angles of rotation <p x , <p y , <p z , about
206
RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
[CHAP. 5
three perpendicular axes, required to bring the body to this orientation.
However, a little experimenting with a solid body will convince anyone
that no suitable coordinates of this sort exist. Consider, for example,
the position specified by <p x = 90°, <p y = 90°, <p z = 0. If a body is first
rotated 90° about the x-axis, and then 90° about the 2/-axis, the final posi-
tion will be found to be different from that resulting from a 90° rotation
about the y-axis followed by a 90° rotation about the x-axis. It turns out
that no simple symmetric set of coordinates can be found to describe the
orientation of a body, analogous to the coordinates x, y, z which locate the
position of a point in space. We therefore postpone to Chapter 11 the
treatment of the rather difficult problem of the rotation of a body around
a point. We shall discuss here only the simple problem of rotation about
a fixed axis.
5-2 Rotation about an axis. It requires only one coordinate to specify
the orientation of a body which is free to rotate only about a fixed axis.
Let the fixed axis be taken as the z-axis, and let a line OA in the body,
through the axis and lying in (or parallel to) the xy-plane, be chosen. We
fix the position of the body by specifying the angle between the line OA
fixed in the body and the x-axis. Choosing cylindrical coordinates to
locate each particle in the body, we now compute the total angular momen-
tum about the z-axis. (See Fig. 5-2.) We shall write r, instead of p» to
represent the distance of particle ro 8 - from the z-axis, in order to avoid con-
fusion with the density p :
L = ]C m » r ^;- (5-6)
i
Let fa be the angle between the direction of the line OA in the body and
the direction of the radius from the z-axis to the particle rm. Then, for a
Fig. 5-2. Coordinates of a particle in a rigid body.
5-2] ROTATION ABOUT AN AXIS 207
rigid body, ft is constant, and
<Pi = + Pi, (5-7)
<Pi = e. (5-8)
Substituting in Eq. (5-6), we have
L = 2^ m i r i^
= (E rmr f\
= 1.6, (5-9)
where
The quantity 7« is a constant for a given body rotating about a given axis,
and is called the moment of inertia about that axis. We may also express
I t as an integral over the body:
fffpr 2 dV. (5-11)
(body)
It is sometimes convenient to introduce the radius of gyration k z denned by
the equation
Mh\ = /,; (5-12)
that is, k z is a radius such that if all the mass of the body were situated a
distance k z from the axis, its moment of inertia would be I„.
Using Eq. (5-9), we may write the component of Eq. (5-5) along the
axis of rotation in the form
^ = 1,9 = N z , (5-13)
where N z is the total external torque about the axis. Equation (5-13) is
the equation of motion for rotation of a rigid body about a fixed axis. It
has the same form as Eq. (2-1) for the motion of a particle along a straight
line. The problem of rotation of a body about a fixed axis is therefore
equivalent to the problem treated in Chapter 2. All methods and results
of Chapter 2 can be extended directly to the present problem according to
the following scheme of analogy:
208
RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
[CHAP. 5
Rectilinear motion
Rotation about a fixed axis
position:
X
angular position :
velocity :
v =
X
angular velocity: w = 6
acceleration :
a =
X
angular acceleration : a = S
force :
F
torque : N z
mass:
m
moment of inertia: I z
potential energy:
potential energy:
V(x) = -
7*
F(x) dx
V(d) = - f" N z (e) de
F(x) = -
dV
dx
»m = - W
kinetic energy:
T =
%mx 2
kinetic energy: T = J/ 2 d
linear momentum
V =
mx
angular momentum : L = I z 6
The only mathematical difference between the two problems is that the
moment of inertia I z depends upon the location of the axis in the body,
while the mass of a body does not depend on its position or on its motion.
This does not affect the treatment of rotation about a single fixed axis.
The rotational potential and kinetic energies defined by the equations,
V{6) = - f N z {6) de,
N z
T
_ dV
de '
iij 2 ,
(5-14)
(5-15)
(5-16)
are not merely analogous to the corresponding quantities defined by Eqs.
(2-41), (2-47), and (2-5) for linear motion. They are, in fact, equal to the
potential and kinetic energies, defined in Chapters 2 and 4, of the system
of particles making up the rigid body. The potential energy defined by
Eq. (5-14), for example, is the work done against the forces whose torque
is N z , when the body is rotated through the angle — e . The kinetic
energy defined by Eq. (5-16) is just the sum of the ordinary kinetic energies
of motion of the particles making up the body. The proof of these state-
ments is left as an exercise.
5-3 The simple pendulum. As an example of the treatment of rota-
tional motion, we consider the motion of a simple pendulum, consisting of
a mass m suspended from a fixed point by a string or weightless rigid rod
5-3]
THE SIMPLE PENDULUM
209
Fig. 5-3. The simple pendulum.
of length I. If a string supports the mass m, we must suppose that it
remains taut, so that the distance I from m to remains constant; other-
wise we cannot treat the system as a rigid one. We consider only motions
of the pendulum in one vertical plane, in order to be able to apply the
simple theory of motion about a single fixed axis through 0. We then
have (Fig. 5-3)
I z = ml 2 , (5-17)
N z = — mgl sin 6,
(5-18)
where the z-axis is an axis through perpendicular to the plane in which
the pendulum is swinging. The torque is taken as negative, since it acts
in such a direction as to decrease the angle 6. Substituting in the equation
of motion (5-13), we find
(5-19)
sin 6.
This equation is not easy to solve. If, however, we consider only small
oscillations of the pendulum (say <3C tt/2), then sin = 0, and we can
write
6 + | 6 = 0. (5-20)
This is of the same form as Eq. (2-84) for the harmonic oscillator. Its
solution is
= k cos (cot + 0), (5-21)
where
(5-22)
= (!)'".
and k and are arbitrary constants which determine the amplitude and
phase of the oscillation. Notice that the frequency of oscillation is
210
RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
[CHAP. 5
independent of the amplitude, provided the amplitude is small enough so
that Eq. (5-20) is a good approximation. This is the basis for the use
of a pendulum to regulate the speed of a clock.
We can treat the problem of motion at large amplitudes by means of
the energy integral. The potential energy associated with the torque
given by Eq. (5-18) is
V(0) = — / —mgl sin dd
J'*
= —mgl cos 0,
(5-23)
where we have taken a — ir/2 for convenience. We could have written
down V(9) right away as the gravitational potential energy of a mass m,
referred to the horizontal plane through as the level of zero potential
energy. The energy integral is
%ml 2 6 2 — mgl cos = E.
(5-24)
We could prove that E is constant from the equation of motion (5-13),
but we need not, since the analogy described in the preceding section
guarantees that all theorems for one-dimensional linear motion will hold in
their analogous forms for rotational motion about an axis. The potential
energy V(8) is plotted in Fig. 5-4. We see that for — mgl < E < mgl, the
motion is an oscillating one, becoming simple harmonic motion for E
slightly greater than —mgl. For E > mgl, the motion is nonoscillatory;
steadily increases or steadily decreases, with oscillating between a maxi-
mum and minimum value. Physically, when E > mgl, the pendulum has
enough energy to swing around in a complete circle. (In this case, of
course, the mass must be held by a rigid rod instead of a string, unless 6 is
very large.) This motion is still a periodic one, the pendulum making
one complete revolution each time increases or decreases by 2ir. In
Fig. 5-4. Potential energy for simple pendulum.
5-3] THE SIMPLE PENDULUM 211
either case, the attempt to solve Eq. (5-24) for leads to the equation
I 9 dO (2gY»
J, (E/mgl + cos $)H» ~\l/ t - {d lb)
The integral on the left must be evaluated in terms of elliptic functions.
The period of the motion can be obtained by integrating between appro-
priate limits. When the motion is oscillatory {E < mgl), the maximum
value k of is given, according to Eq. (5-24), by
E = — mgl cos k. (5-26)
Equation (5-25) becomes, in this case,
f de (2 9 y>
J, (COS - COS K)l/2 _ \ I ) h & Zl)
which can also be written
/
*— -»(i) u \ M
, [sin 2 (k/2) - sin 2 (tf/2)]i/» \l
The angle oscillates between the limits ±k. We now introduce a new
variable <p which runs from to 27T for one cycle of oscillation of 0:
where
sin 8/2 1 . ._ on .
Sin * = ihW2 = a Sin 2' (5 - 29)
a = sin | • (5-30)
With these substitutions, Eq. (5-28) can be written
(fa (g
[
Jo
* - / M/2
o (1 - a 2 sin 2 v )i/2 \l
t, (5-31)
where we have taken O = 0, for convenience. The integral is now in a
standard form for elliptic integrals. When a is small, the integrand can be
expanded in a power series in a 2 :
L
[1 + £a 2 sin 2 <p + ■ ■ ■] d v = (jj t. (5-32)
This can be integrated term by term:
<P + ia 2 (2*> - sin2<p) -] = (j j t. (5-33)
212
BIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
The period of the motion is obtained by setting <p = 2ir :
—©■V*+-)
[chap. 5
(5-34)
Thus as the amplitude of oscillation becomes large, the period becomes
slightly longer than for small oscillations, a prediction which is readily
verified experimentally by setting up two pendulums of equal length and
setting them to swinging at unequal amplitudes. Equation (5-33) can be
solved approximately for <p by successive approximations, and the result
substituted in Eq. (5-29), which can be solved for 8 by successive approxi-
mations. The result, to a second approximation, is
where
= (k + j^) sin o>'t + ^ sin 3c/ 1, (5-35)
"-*-(dVS+")" <->
If we neglect terms in /c 2 and k 3 , this solution agrees with Eq. (5-21). At
larger amplitudes in second approximation, the frequency is slightly lower
than at small amplitudes, and the motion of 8 contains a small third har-
monic term.
5-4 The compound pendulum. A rigid body suspended and free to
swing about an axis is called a compound pendulum. We assume that the
axis does not pass through the center of mass, and we specify the position
of the body by the angle 8 between a vertical line and a perpendicular line
drawn from a point on the axis, through the center of mass G (Fig. 5-5).
In order to compute the total torque exerted by gravity, we anticipate a
Fig. 5-5. The compound pendulum.
5-4] THE COMPOUND PENDULUM 213
theorem, to be proved later, that the total torque is the same as if the
total gravitational force were applied at the center of mass G. We then
have, using Eqs. (5-12) and (5-13),
Mk%e = —Mgh sin 6, (5-37)
where h is the distance OG. This equation is the same as Eq. (5-19) for a
simple pendulum of length I, if we take
* = f- (5-38)
The point 0' a distance I from along the line through the center of
mass G is called the center of oscillation. If all the mass M were at 0',
the motion of the pendulum would be the same as its actual motion, for
any given initial conditions. If the distance O'G is h', we have
I = h + h', (5-39)
hh' = k% — h 2 . (5-40)
It will be shown in the next section that the moment of inertia about
any axis equals the moment of inertia about a parallel axis through the
center of mass G plus Mh 2 , where h is the distance from the axis to G.
Let kg be the radius of gyration about G. We then have
k 2 = k% + h 2 , (6-41)
so that Eq. (5-40) becomes
hh' = k%. (5-42)
Since this equation is symmetrical in h and hi , we conclude that if the
body were suspended about a parallel axis through 0', the center of oscilla-
tion would be at 0. The acceleration g of gravity can be measured very
accurately by measuring the period of small oscillations of a pendulum and
using Eq. (5-22). If a compound pendulum is used, the radius of gyration
must be known, or the period measured about two axes, preferably 0, 0',
so that the radius of gyration can be eliminated from the equations.
Consider a rigid body suspended from an axis about which it is free to
move. Let it be struck a blow at a point 0' a distance I from the axis,
the direction of the blow being perpendicular to the line 00' from the
axis to 0'. Place 0' so that the line 00' passes through the center of mass
G, and let h, h' be the distances OG, O'G (Fig. 5-6). The impulse delivered
at the point 0' by the force F' during the blow is
V = />
dt. (5-43)
214
RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
) '"
h
Q»
1
h'
0'
I »
/ r
Fig. 5-6. Rigid body pivoted at and struck a blow at 0'.
At the instant the blow is struck, a force F will, in general, have to be
exerted on the body at the point on the axis in order to keep fixed.
The impulse delivered to the body at is
-/,
dt.
(5-44)
An equal and opposite impulse — / is delivered by the body to the support
at 0. The momentum theorem for the component P of linear momentum
of the body in the direction of F is:
f = J t Wh8) = F + F',
(5-45)
where 6 is the angular velocity of the body about 0. From this we have,
for the momentum just after the blow,
MU = J + J',
(5-16)
assuming that the body is initially at rest. The conservation theorem of
angular momentum about is:
§ =±(Mkle) = F>l. (H7)
Integrating, we have, for the angular momentum just after the blow,
Mkl6 = J'l. (5-48)
We eliminate 6 between Eqs. (5-46) and (5-48) :
= h% (l + j) • (5-49)
hi
5-5] COMPUTATION OF CENTERS OF MASS AND MOMENTS OF INERTIA 215
We now ask for the condition that no impulsive force be exerted on the
axis at at the instant of the blow, i.e., J = 0:
hi = h%. (5-50)
This equation is identical with Eq. (5-38) and may also be expressed in
the symmetrical form [Eq. (5-42)]
hh' = k%. (5-51)
The point 0' at which a blow must be struck in order that no impulse be
felt at the point is called the center of percussion relative to 0. We see
that the center of percussion is the same as the center of oscillation relative
to 0, and that is the center of percussion relative to 0'. If the body is
unsupported, and is struck at 0', its initial motion will be a rotation about
0. For example, a batter tries to hit a baseball at the center of percussion
relative to his hands. If the ball hits very far from the center of percus-
sion, the blow is transmitted to his hands by the bat.
5-5 Computation of centers of mass and moments of inertia. We have
given in Section 4-1 the following definition of center of mass for a system
of particles :
R = m E «**• ( 5 ~ 52 )
For a solid body, the sum may be expressed as an integral:
R = jjfff P TdV, (5-53)
or, in component form,
X = ±fffpxdV, (5-54)
Y = IfIfI pydV > (5_55)
Z = ±fffpzdV. (5-56)
The integrals can be extended either over the volume of the body, or over
all space, since p = outside the body. These equations define a point G
of the body whose coordinates are (X, Y, Z). We should first prove that
the point G thus defined is independent of the choice of coordinate system.
Since Eq. (5-52) or (5-53) is in vector form, and makes no reference to
any particular set of axes, the definition of G certainly does not depend on
216
RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
[CHAP. 5
Fig. 5-7. Location of center of mass relative to two different origins.
any particular choice of directions for the axes. We should prove, how-
ever, that G is independent also of the choice of origin. Consider a system
of particles, and let any particle ra; be located by vectors r t - and x\ with
respect to any two origins and 0'. If a is the vector from to 0', the
relation between r; and r£ is (Fig. 5-7)
ft + a.
(5-57)
The centers of mass G, G' with respect to 0, 0' are located by the vectors
R and R', where R' is defined by
R ' = m £ m ^-
i
Using Eq. (5-57), we can rewrite Eq. (5-58) :
i
= MZ m ^-M a E m »
i i
= R - a.
(5-58)
(5-59)
Thus R and R' are vectors locating the same point with respect to and 0'
so that G and G' are the same point.
General theorems like the one above can be proved either for a system of
particles or for a body described by a density p. Whichever point of view
is adopted in any proof, a parallel proof can always be given from the
other point of view.
Much of the labor involved in the calculation of the position of the
center of mass from Eqs. (5-54), (5-55), (5-56) can often be avoided by
the use of certain laborsaving theorems, including the theorem proved
5-5] COMPUTATION OF CENTERS OF MASS AND MOMENTS OF INERTIA 217
above which allows us a free choice of coordinate axes and origin. We
have first the following theorem regarding symmetrical bodies:
Theorem. If a body is symmetrical with respect to a plane, its
center of mass lies in that plane. (5-60)
When we say a body is symmetrical with respect to a plane, we mean that
for every particle on one side of the plane there is a particle of equal mass
located at its mirror image in the plane. For a continuously distributed
mass, we mean that the density at any point equals the density at its
mirror image in the plane. Choose the origin in the plane of symmetry,
and let the plane of symmetry be the xy-plaue. Then in computing Z from
Eq. (5-56) [or (5-52)], for each volume element (or particle) at a point
(x, y, z) above the xy-plane, there is, by symmetry, a volume element of
equal mass at the point (x, y, —z) below the zy-plane, and the contribu-
tions of these two elements to the integral in Eq. (5-56) will cancel. Hence
Z = 0, and the center of mass lies in the xy-p\a,ne. This proves Theorem
(5-60). The theorem has a number of obvious corollaries:
If a body is symmetrical in two planes, its center of mass lies on
their line of intersection. (5-61)
// a body is symmetrical about an axis, its center of mass lies on
that axis. (5-62)
If a body is symmetrical in three planes with one common point,
that point is its center of mass. (5-63)
7/ a body has spherical symmetry about a point (i.e., if the density
depends only on the distance from that point), that point is its
center of mass. (5-64)
These theorems enable us to locate the center of mass immediately in some
cases, and to reduce the problem to a computation of only one or two
coordinates of the center of mass in other cases. One should be on the
lookout for symmetries, and use them to simplify the problem. Other
cases not included in these theorems will occur (e.g., the parallelepiped),
where it will be evident that certain integrals will be equal or will cancel,
and the center of mass can be located without computing them.
Another theorem which often simplifies the location of the center of
mass is that if a body is composed of two or more parts whose centers of
mass are known, then the center of mass of the composite body can be
computed by regarding its component parts as single particles located at
their respective centers of mass. Let a body, or system of particles, be
composed of n parts of masses M\, . . . , M n . Let any part M& be composed
of Nk particles of masses muu ■ ■ ■ , mkN k , located at the points r^, . . . , tkN k -
218 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
Then the center of mass of the part Mu is located at the point
and
i _ k _
M k = X m k i.
1=1
(5-65)
(5-66)
The center of mass of the entire body is located at the point
where
k=i i=i
h=l 1=1
By Eq. (5-65), Eq. (5-67) becomes
M
k=i
and by Eq. (5-66), Eq. (5-68) becomes
M = J) M k .
(5-67)
(5-68)
(5-69)
(5-70)
A-=l
Equations (5-69) and (5-70) are the mathematical statement of the theo-
rem to be proved.
5-5] COMPUTATION OF CENTERS OF MASS AND MOMENTS OF INERTIA 219
As an example, let us consider a uniform rectangular block with a cylin-
drical hole drilled out, as shown in Fig. 5-8. By the symmetry about the
two vertical planes bisecting the block parallel to its sides, we conclude
that the center of mass lies along the vertical line AB through the centers
of the top and bottom faces. Let the center of mass of the block lie a
distance Z below A, and let the density of the block be p. If the hole were
not cut out, the mass of the block would be 6 cm X 4 cm X 10 cm X p,
and its center of mass would be at the midpoint of AB, 5 cm from A. The
mass of the material drilled out is ir cm 2 X 6 cm X p, and its center of
mass, before it was removed, was on AB, 2 cm below A. Hence the
theorem (5-69) above allows us to write
(6 cm X 4 cm X 10 cm X p) X 5 cm = (7r cm 2 X 6 cm X p) X 2 cm
+ 6 cm X (4 cm X 10 cm — tt cm 2 ) X p X Z.
The solution for Z is
6X4X10X5 — 7TX6X2
Z = 6X(4X10-tt) Cm -
As a second example, we locate the center of mass of a hemisphere of
radius a. By symmetry, if the density is uniform, the center of mass lies
on the axis of symmetry, which we take as the z-axis. We have then to
(a)
(b)
(c) (d)
Fig. 5-9. Methods of integrating over a hemisphere.
220 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
compute only the integral in Eq. (5-56). The integral can be set up in
rectangular, cylindrical, or spherical coordinates (Fig. 5-9) :
ra /-(a 2 -* 2 ) 1 ' 2 r^-z 2 -^) 1 ' 2
Rectangular: Z = £ J ^ j^^^ J^^^u* >> z dx * *■
ra r2* r (a 2 -* 2 ) 1 ' 2
Cylindrical : Z = -jj / / / pzr dr dip dz.
Ml y 2== o J <p=o Jr=o
ra r*l2 r2r
Spherical : Z = -^ I J I (pr cos 0)r 2 sin dr dO dip.
M. J r== o Je=,o J v=o
Any one of these expressions can be used to evaluate Z for any density
distribution. If p is uniform, we can also build up the hemisphere out of
rings or disks and save one or two integrations. For example, building up
the hemisphere out of disks perpendicular to the z-axis (this is equivalent
to carrying out the integration over r and <p in cylindrical coordinates), we
can write
z ~h/,.o z '" ( ° 2 "* 2) *
-GspsXt*)-* (5 " 71 >
where the integrand is zp times the volume of a disk of thickness dz, radius
(a 2 - z 2 ) 1 ' 2 .
When the density p is uniform, the center of mass of a body depends
only on its geometrical shape, and is given by
K = ±fffrdV. (5-72)
v
The point G whose coordinate R is given by Eq. (5-72) is called the cen-
troid of the volume V. If we replace the volume V by an area A or curve C
in space, we obtain formulas for the centroid of an area or of a curve:
R = jffrdA, (5-73)
A
R = i J r da, (5-74)
where s is the length of the curve C. The following two theorems, due to
Pappus, relate the centroid of an area or curve to the volume or area swept
out by it when it is rotated about an axis:
5-51 COMPUTATION OF CEN'TERS OF MASS AND MOMENTS OF INERTIA 221
Fig. 5-10. Pappus' first theorem.
Fig. 5-11. Sphere formed by rota-
ting a semicircle.
Theorem 1. If a plane curve rotates about an axis in its own
plane which does not intersect it, the area of the surface of revolu-
tion which it generates is equal to the length of the curve multiplied
by the length of the path of its centroid, (5-75)
Theorem 2. If a plane area rotates about an axis in its own plane
which does not intersect it, the volume generated is equal to the area
times the length of the path of its centroid. (Z
-76)
The proof of Theorem 1 is very simple, with the notation indicated in
Fig. 5-10:
A = f 2iryds= 2tt f y ds = 2ttYs, (5-77)
where Y is the y-co ordinate of the centroid of the curve C, and s is its
length. The proof of Theorem 2 is similar and is left to the reader. These
theorems may be used to determine areas and volumes of figures sym-
metrical about an axis when the centroids of the generating curves or
areas are known, and conversely. We locate, for example, the position of
the center of mass of a uniform semicircular disk of radius a, using Pappus'
second theorem. If the disk is rotated about its diameter, the volume of
the sphere generated, by Pappus' theorem (Fig. 5-11), is
fra> = (^) (2*7),
from which we obtain
Y m
ia
3tt'
(5-78)
The moment of inertia / of a body about an axis is defined by Eq, (5-1 0) :
/ = £ rmrl (5-79)
222 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
P
Fig. 5-12. Location of point P with respect to points and G.
or
1 = fffpr 2 dV,
(5-80)
where r is the distance from each point or particle of the body to the given
axis. We first prove several laborsaving theorems regarding moments
of inertia :
Parallel Axis Theorem. The moment of inertia of a body
about any given axis is the moment of inertia about a parallel axis
through the center of mass, plus the moment of inertia about the
given axis if all the mass of the body were located at the center of
mass. (5-81)
To prove this theorem, let Io be the moment of inertia about a z-axis
through the point 0, and let Ig be the moment of inertia about a parallel
axis through the center of mass G. Let r and r' be the vectors to any
point P in the body, from and G, respectively, and let R be the vector
from to G. The components of these vectors will be designated by
(x, y, z), (x f , y', z'), and (X, Y, Z). Then, since (Fig. 5-12)
r = r' + R,
we see that
x 2 + y 2 = (x' + X) 2 + (y' + Y) 2
= x' 2 + y' 2 + X 2 + Y 2 + 2Xx' + 2Yy',
so that the moment of inertia Io is
Io = fff(x 2 + y 2 )pdV
= fff(x' 2 + y' 2 )p dV + (X 2 + Y 2 ) Jffp dV + 2X fjfx'p dV
+ 2Yfffy'pdV. (5-82)
5-5] COMPUTATION OF CENTERS OF MASS AND MOMENTS OF INERTIA 223
The first integral is la, and the integral in the second term is the total
mass M of the body. The integrals in the last two terms are the same as
the integrals occurring in Eqs. (5-54) and (5-55), and define the x- and
^/-coordinates of the center of mass relative to G. Since G is the center of
mass, these integrals are zero, and we have
lo = Ig + M(X 2 + Y 2 ). (5-83)
This is the mathematical statement of the Parallel Axis Theorem. If we
know the moment of inertia about any axis, and can locate the center of
mass, we can use this theorem to determine the moment of inertia about
any other parallel axis.
The moment of inertia of a composite body about any axis may be
found by adding the moments of inertia of its parts about the same axis, a
statement which is obvious from the definition of moment of inertia. This
fact can be put to use in the same way as the analogous result for the
center of mass of a composite body.
A body whose mass is concentrated in a single plane is called a plane
lamina. We have the following theorem for a plane lamina:
Perpendicular Axis Theorem. The sum of the moments of
inertia of a plane lamina about any two perpendicular axes in the
plane of the lamina is equal to the moment of inertia about an axis
through their point of intersection perpendicular to the lamina. (5-84)
The proof of this theorem is very simple. Consider any particle of mass m
in the xy-pl&ne. Its moments of inertia about the x- and y-axes are
I x = my 2 , I y = mx 2 . (5-85)
Adding these, we have the moment of inertia of m about the z-axis:
h + I v = m(x 2 + y 2 ) = /*. (5-86)
Since the moment of inertia of any lamina in the rcy-plane is the sum of
the moments of inertia of the particles of which it is composed, we have
theorem (5-84).
We illustrate these theorems by finding the moments of inertia of a
uniform circular ring of radius a, mass M, lying in the xy-p\&ne (Fig. 5-13).
The moment of inertia about a z-axis perpendicular to the plane of the ring
through its center is easily computed:
7, = Ma 2 . (5-87)
The moments I x and I y are evidently equal, and we have, therefore, by
theorem (5-84),
I x = \l t = %Ma 2 . (5-88)
224 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
Fig. 5-13. A ring of radius o.
Fig. 5-14. Finding the moment of
inertia of a disk.
Fig. 5-15. Finding the moment of inertia of a solid sphere.
The moment of inertia about an axis A tangent to the ring is, by the Parallel
Axis Theorem,
I A = l x + Ma 2 = §Ma 2 . (5-89)
The moment of inertia of a solid body can be set up in whatever coordi-
nate system may be convenient for the problem at hand. If the body is
uniform and of simple shape, its moment of inertia can be computed by
considering it as built up out of rods, rings, disks, etc. For example, the
moment of inertia of a circular disk about an axis perpendicular to it
through its center can be found by regarding the disk as made up of rings
(Fig. 5-14) and using Eq. (5-87) :
pa
Jo
r p2irr dr
TOjO
\Ma 2 .
(5-90)
The moment of inertia of a solid sphere can be calculated from Eq. (5-90)
by regarding the sphere as made up of disks (Fig. 5-15) :
I = f Sl^li {(ma 2 sin 2 0) d(a cos B) = ^f = \Ma 2 . (5-91)
5-6] STATICS OF RIGID BODIES 225
A body with a piece cut out can be treated by setting its moment of
inertia equal to the moment of inertia of the original body minus the
moment of inertia of the piece cut out, all moments being taken, of course,
about the same axis.
5-6 Statics of rigid bodies. The equations of motion of a rigid body
are Eqs. (5-4) and (5-5) :
MR = J2 F * (5-92)
i
^f=Z N t-o. (5-93)
i
Equation (5-92) determines the motion of the center of mass, located by
the vector R, in terms of the sum of all external forces acting on the body.
Equation (5-93) determines the rotational motion about a point 0, which
may be the center of mass or a point fixed in space, in terms of the total
external torque about the point 0. Thus if the total external force acting
on a rigid body and the total external torque about a suitable point are
given, its motion is determined. This would not be true if the body were
not rigid, since then it would be deformed by the external forces in a man-
ner depending on the particular points at which they are applied. Since
we are concerned only with external forces throughout this section, we
may omit the superscript e. It is only necessary to give the total torque
about any one point 0, since the torque about any other point 0' can then
be found from the following formula:
J2 Ha,' = X) N i0 + (r G - to') X £ F f , (5-94)
i i i
where io, r < are vectors drawn to the points 0, 0' from any convenient
origin. That is, the total torque about 0' is the total torque about plus
the torque about 0' if the total force were acting at 0. The proof of
Eq. (5-94) is very simple. Let r» be the vector from the origin to the point
at which Fj acts. Then
J2 N i0 ' = ]£ (ii - to-) X F s -
i i
= S (fi — to + TO — To') X F t -
i
= S ( r < - f o) X F»- + X) (To - T ') X Fi
i i
= £ N i0 + (r - to') X J2 *<•
226 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
P i?
0"
Fig. 5-16. The torque is independent of where along its line of action a force acts.
If, in particular, a rigid body is at rest, the left members of Eqs. (5-92)
and (5-93) are zero, and we have
i
J2 N * = °- ( 5 ~ 96 )
i
These are the conditions to be satisfied by the external forces and torques
in order for a rigid body to be in equilibrium. They are not sufficient to
guarantee that the body is at rest, for it might still be in uniform transla-
tional and rotational motion, but if the body is initially at rest, it will
remain at rest when these conditions are satisfied. It is sufficient for the
total torque in Eq. (5-96) to be zero about any point, since then, by
Eq. (5-94), it will be zero also about every other point if Eq. (5-95) holds.
In computing the torque due to a force F, it is necessary to know not
only the vector F (magnitude and direction), but also the point P of
the body at which the force acts. But if we draw a line through P in
the direction of F, then if F acts at any other point P' of this line, its
torque will be the same, since, from the definition of the cross product,
it can be seen (Fig. 5-16) that
r P X F = r P > X F. (5-97)
(The areas of the parallelograms involved are equal.) The line through P
in the direction of F is called the line of action of the force. It is often
convenient in computing torques to remember that the force may be con-
sidered to act anywhere along its line of action. A distinction is some-
times made in this connection between "free" and "sliding" vectors, the
force being a "sliding" vector. The terminology is likely to prove confus-
ing, however, since as far as the motion of the center of mass is concerned
[Eq. (5-92)], the force is a "free" vector, i.e., may act anywhere, whereas
in computing torques, the force is a "sliding" vector, and for a nonrigid
body, each force must be localized at the point where it acts. It is better
to define vector, as we defined it in Section 3-1, as a quantity having
magnitude and direction, without reference to any particular location in
space. Then, in the case of force, we need for some purposes to specify not
5-6] STATICS OF RIGID BODIES 227
Fig. 5-17. A single force C whose torque is the sum of the torques of A and B.
only the force vector F itself, but in addition the point or line on which
the force acts.
A theorem due to Varignon states that if C = A + B, then the moment
of C about any point equals the sum of the moments of A and B, provided
A, B, and C act at the same point. The theorem is an immediate conse-
quence of the vector identity given by Eq. (3-27) :
rxC = rxA + rxB, if C = A + B. (5-98)
This theorem allows us to compute the torque due to a force by adding
the torques due to its components. Combining Varignon's theorem with
the result of the preceding paragraph, we may reduce the torque due to
two forces A, B acting in a plane, as shown in Fig. 5-17, to the torque
due to the single force C, since both A and B may be considered to act at
the intersection of their lines of action, and Eq. (5-98) then allows us to
add them. We could now add C similarly to any third force acting in the
plane. This process can be continued so long as the lines of action of the
forces being added are not parallel, and is related to a more general theorem
regarding forces in a plane to be proved below.
Since, for a rigid body, the motion is determined by the total force and
total torque, we shall call two systems of forces acting on a rigid body
equivalent if they give the same total force, and the same total torque
about every point. In view of Eq. (5-94), two systems of forces are then
equivalent if they give the same total force, and the same total torque
about any single point. It is of interest to know, for any system of forces,
what is the simplest system of forces equivalent to it.
If a system of forces F t - acting at points r t - is equivalent to a single force F
acting at a point r, then the force F acting at r is said to be the resultant of
the system of forces F,-. If F is the resultant of the system of forces F;,
then we must have
F = Z) Fi, (5-99)
(r-ro)xF=^ (r« - r ) X F if (5-100)
228 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
where r is any point about which moments are taken. By Eq. (5-94), if
Eq. (5-99) holds, and Eq. (5-100) holds for any point r , it holds for every
point r . The force — F acting at r is called the equilibrant of the system;
if the equilibrant is added to the system of forces, the conditions for equi-
librium are satisfied.
An example of a system of forces having a resultant is the system of
gravitational forces acting on a body near the surface of the earth. We
shall show that the resultant in this case acts at the center of mass. Let
the acceleration of gravity be g. Then the force acting on a particle m< is
F< = m ig . (5-101)
The total force is
F = £ »« = M &> ( 5 - 102 )
i
where M is the total mass. The total torque about any point is, with
as origin,
^2 n j0 = J2 ( r * x m *s)
i i
x g
= te m * r »)
= MR x g
= R X Ig, (5-103)
where R is the vector from to the center of mass. Thus the total torque
is given by the force Mg acting at the center of mass. Because of this
result, the center of mass is also called the center of gravity. We shall see
in the next chapter that, in general, this result holds only in a uniform
gravitational field, i.e., when g is the same at all points of the body. If
the system of forces acting on a rigid body has a resultant, the forces may
be replaced by this resultant in determining the motion of the body.
A system of forces whose sum is zero is called a couple:
2 F * = °- ( 5 " 104 )
i
A couple evidently has no resultant, except in the trivial case where the
total torque is zero also, in which case the resultant force is zero. By
Eqs. (5-94) and (5-104), a couple exerts the same total torque about every
point :
X) N i0 < = X) N <°- ( 5 - 105 )
5-6] STATICS OF RIGID BODIES 229
P
p,
Fig. 5-18. A simple couple.
Thus a couple is characterized by a single vector, the total torque, and
all couples with the same total torque are equivalent. The simplest sys-
tem equivalent to any given couple, if we exclude the trivial case where
the total torque is zero, is a pair of equal and opposite forces F, — F, acting
at points P, P' separated by a vector r (Fig. 5-18) such that
J2 Nio = r X F. (5-106)
i
Equation (5-106) states that the moment of the given couple about
equals the moment of the couple (F, — F) about P'; the two systems
are therefore equivalent, since the point about which the moment of a
couple is computed is immaterial. The force F and the points P and P'
are by no means uniquely determined. Since only the cross product r X F
is determined by Eq. (5-106), we can choose P arbitrarily; we can choose
the vector F arbitrarily except that it must lie in the plane perpendicular to
the total torque; and we can then choose r as any vector lying in the same
plane and determining with F a parallelogram whose area is the magnitude
of the total torque.
The problem of finding the simplest system equivalent to any given sys-
tem of forces is solved by the following theorems :
Theorem I. Every system of forces is equivalent to a single force
through an arbitrary point, plus a couple (either or both of which
may be zero). (5-107)
To prove this, we show how to find the equivalent single force and couple.
Let the arbitrary point P be chosen, let the sum of all the forces in the sys-
tem be F, and let their total torque about the point P be N. Then, if we
let the single force F act at P, and add a couple whose torque is N, we have
a system equivalent to the original system. Since the couple can be com-
posed of two forces, one of which may be allowed to act at an arbitrary
point, we may let one force of the couple act at the point P, and add it to
F to get a single force acting at P plus the other force of the couple. This
proves
Theorem II. Any system of forces can be reduced to an equiva-
lent system which contains at most two forces. (5-108)
230 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
The following theorem can be proved in two ways:
Theorem III. A single nonzero force and a couple in the same
plane (i.e., such that the torque vector of the couple is perpendicular
to the single force) have a resultant and, conversely, a single force
is equivalent to an equal force through any arbitrary point, plus a
couple. (5-109)
Since a couple with torque N is equivalent to a pair of equal and opposite
forces, F, — F, where F may be chosen arbitrarily in the plane perpendicular
to N, we may always choose F equal to the single force mentioned in the
theorem. Furthermore, we may choose the point of action of F arbitrarily.
Given a single nonzero force F acting at P, and a couple, we form a couple
(F, — F) equivalent to the given couple, and let — F act at P; F and — F
then cancel at P, and the remaining force F of the couple is the single re-
sultant. The converse can be proved by a similar argument.
The other method of proof is as follows. Let the given force F act at a
point P, and let the total torque of the couple be N. Then the torque of
the system about the point P is N. We take any vector r, in the plane
perpendicular to N, which forms with F a parallelogram of area N, and
let P' be the point displaced from P by the vector r. If the single force F
acts at P', the torque about P will then be N, and hence this single force is
equivalent to the original force F acting at P plus the couple. We can
combine Theorems I and III to obtain
Theorem IV. Every system of forces is equivalent to a single
force plus a couple whose torque is parallel to the single force.
(Or, alternatively, every system of forces is equivalent to a couple
plus a single force perpendicular to the plane of the couple.) (5-110)
To prove this, we use Theorem I to reduce any system to a single force
plus a couple, and use Theorem III to eliminate any component of the
couple torque perpendicular to the single force. The point of application
of the single force mentioned in Theorem IV is no longer arbitrary, as its
line of action will be fixed when we apply Theorem III. Either the single
force or the couple may vanish in special cases. For a system of forces in
a plane, all torques about any point in the plane are perpendicular to the
plane. Hence Theorem IV reduces to
Theorem V. Any system of forces in a plane has a resultant,
unless it is a couple. (5-111)
In practice, the reduction of a complicated system of forces to a simpler
system is a problem whose simplest solution is usually obtained by an
ingenious application of the various theorems and techniques mentioned
in this section. One method which always works, and which is often the
5-7] STATICS OF STRUCTURES 231
simplest if the system of forces is very complicated, is to follow the pro-
cedure suggested by the proofs of the above theorems. Find the total
force F by vector addition, and the total torque N about some conven-
iently chosen point P. Then F acting at P, plus a couple of torque N,
together form a system equivalent to the original system. If F is zero, the
original system reduces to a couple. If N is perpendicular to F, the system
has a resultant, which can be found by either of the methods indicated in
the proof of Theorem III. If N is not perpendicular to F, and neither is
zero, then the system has no resultant, and can be reduced to a system of
two forces, as in the derivation of Theorem II, or to a single force and a
couple whose torque is parallel to it, as in Theorem IV. It is a matter of
taste, or of convenience for the purpose at hand, which of these latter re-
ductions is regarded as the simplest. In fact, for determining the motion
of a body, the most convenient reduction is certainly just the reduction
given by Theorem I, with the arbitrary point taken as the center of mass.
5-7 Statics of structures. The determination of the forces acting at
various points in a solid structure is a problem of utmost importance in
all phases of mechanical engineering. There are two principal reasons for
wanting to know these forces. First, the engineer must be sure that the
materials and construction are such as will withstand the forces which will
be acting, without breaking or crushing, and usually without suffering
permanent deformation. Second, since no construction materials are
really rigid, but deform elastically and sometimes plastically when subject
to forces, it is necessary to calculate the amount of this deformation, and
to take it into account, if it is significant, in designing the structure. When
deformation or breaking of a structure is under consideration, the struc-
ture obviously cannot be regarded as a rigid body, and we are interested
in the actual system of forces acting on and in the structure. Theorems
regarding equivalent systems of forces are not of direct interest in such
problems, but are often useful as tools in analyzing parts of the structure
which may, to a sufficient approximation, be regarded as rigid, or in sug-
gesting possible equivalent redistributions of forces which would subject
the structure to less objectionable stresses while maintaining it in equi-
librium.
If a structure is at rest, Eqs. (5-95) and (5-96) are applicable either to
the structure as a whole, or to any part of it. It must be kept in mind that
the forces and torques which are to be included in the sums are those
which are external to and acting on whichever part of the structure is under
consideration. If the structure is moving, the more general equations
(5-92) and (5-93) are applicable. Either pair of vector equations repre-
sents, in general, six component equations, or three if all forces lie in a
single plane. (Why three?) It may be that the structure is so constructed
232
RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
[CHAP. 5
Fig. 5-19. The flagpole problem.
that when certain of the external forces and their points of application are
given, all the internal forces and torques acting on each part of the struc-
ture can be determined by appropriate applications of Eqs. (5-95) and
(5-96) (in the case of a structure at rest). Such a structure is said to be
statically determinate. An elementary example is shown in Fig. 5-19, which
shows a horizontal flagpole AB hinged at point A to a wall and supported
by a cable BC. A force W acts on the pole as shown. When the force W
and the dimensions of the structure are given, it is a simple matter to
apply Eqs. (5-95) and (5-96) to the pole and to calculate the force F x ex-
erted by the cable and the force F 2 acting through the hinge. Many ex-
amples of statically determinate structures are given in any elementary
physics textbook.
Suppose now that the hinge at A in Fig. 5-19 were replaced by a welded
joint, so that the flagpole would support the load even without the cable
BC, provided the joint at A does not break. Then, given only the weight
W, it is evidently impossible to determine the force F x exerted by the
cable; Fi may have any value from zero to a rather large value, depending
on how tightly the cable is drawn up and on how much stress is applied
to the joint at A. Such a structure is said to be statically indeterminate.
A statically indeterminate structure is one in which the forces acting on
its parts are not completely determined by the external forces, but depend
also on the distribution of stresses within the structure. To find the
internal forces in an indeterminate structure, we would need to know the
elastic characteristics of its parts and the precise way in which these parts
are distorted. Such problems are usually far more difficult than problems
involving determinate structures. Many methods of calculating internal
forces in mechanical structures have been developed for application to en-
gineering problems, and some of these are useful in a wide variety of
physical problems.
5-8 Stress and strain. If an imaginary surface cuts through any part
of a solid structure (a rod, string, cable, or beam), then, in general, the
material on one side of this surface will be exerting a force on the material
5-8]
STRESS AND STBAIN
233
F^ii F(- r
(a)
PWiFh
(b)
Fj~r
(C)
F,-w
Fig. 5-20. Stresses in a beam, (a) Compression, (b) Tension, (c) Shear.
on the other side, and conversely, according to Newton's third law. These
internal forces which act across any surface within the solid are called
stresses. The stress is defined as the force per unit area acting across any
given surface in the material. If the material on each side of any surface
pushes on the material on the other side with a force perpendicular to the
surface, the stress is called a compression. If the stress is a pull perpen-
dicular to the surface, it is called a tension. If the force exerted across the
surface is parallel to the surface, it is called a shearing stress. Figure 5-20
illustrates these stresses in the case of a beam. The vector labeled Fj_» r
represents the force exerted by the left half of the beam on the right half,
and the equal and opposite force F r _j is exerted on the material on the
left by the material on the right. A stress at an angle to a surface can be
resolved into a shear component and a tension or compression component.
In the most general case, the stress may act in any direction relative to
the surface, and may depend on the orientation of the surface. The de-
scription of the state of stress of a solid material in the most general
case is rather complicated, and is best accomplished by using the mathe-
matical techniques of tensor algebra to be developed in Chapter 10. We
shall consider here only cases in which either the stress is a pure com-
pression, independent of the orientation of the surface, or in which only
one surface is of interest at any point, so that only a single stress vec-
tor is needed to specify the force per unit area across that surface.
If we consider a small volume AV of any shape in a stressed material,
the material within this volume will be acted on by stress forces exerted
across the surface by the material surrounding it. If the material is not
perfectly rigid, it will be deformed so that the material in the volume AV
may have a different shape and size from that which it would have if there
were no stress. This deformation of a stressed material is called strain.
234 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
The nature and amount of strain depend on the nature and magnitude of
the stresses and on the nature of the material. A suitable definition of
strain, stating how it is to be measured, will have to be made for each kind
of strain. A tension, for example, produces an extension of the material,
and the strain would be defined as the fractional increase in length.
If a wire of length I and cross-sectional area A is stretched to length
I + Al by a force F, the definitions of stress and strain are
stress = F/A, (5-112)
strain = Al/l. (5-113)
It is found experimentally that when the strain is not too large, the stress
is proportional to the strain for solid materials. This is Hooke's law, and
it is true for all kinds of stress and the corresponding strains. It is also
plausible on theoretical grounds for the reasons suggested in the preliminary
discussion in Section 2-7. The ratio of stress to strain is therefore constant
for any given material if the strain is not too large. In the case of exten-
sion of a material in one direction due to tension, this ratio is called Young's
modulus, and is
Y = stress = JL . (5_ n4 )
strain A Al
If a substance is subjected to a pressure increment Ap, the resulting
deformation will be a change in volume, and the strain will be denned by
AV
strain = -^=- ■ (5-115)
The ratio of stress to strain in this case is called the bulk modulus B:
„ stress ApV , .
B = — — — = f==- > (5-116)
strain AK
where the negative sign is introduced in order to make B positive.
In the case of a shearing stress, the stress is again denned by Eq. (5-112),
where F is the force acting across and parallel to the area A. The result-
ing shearing strain consists in a motion of A parallel to itself through a
distance Al, relative to a plane parallel to A at a distance Ax from A (Fig.
5-21). The shearing strain is then denned by
strain = — = tan 0, (5-117)
where 6 is the angle through which a line perpendicular to A is turned as
a result of the shearing strain. The ratio of stress to strain in this case is
5-9]
EQUILIBRIUM OF FLEXIBLE STRINGS AND CABLES
235
Fig. 5-21. Shearing strain.
called the shear modulus n:
n =
stress _ F
strain — A tan
(5-118)
An extensive study of methods of solving problems in statics is outside
the scope of this text. We shall restrict ourselves in the next three sec-
tions to the study of three special types of problems which illustrate the
analysis of a physical system, to determine the forces which act upon its
parts and to determine the effect of these forces in deforming the system.
5-9 Equilibrium of flexible strings and cables. An ideal flexible string
is one which will support no compression or shearing stress, nor any bending
moment, so that the force exerted across any point in the string can only
be a tension directed along the tangent to the string at that point. Chains
and cables used in many structures can be regarded for most purposes as
ideal flexible strings.
Let us first take a very simple problem in which a string of negligible
weight is suspended between two points P and P 2 , and a force F x acts at a
point Pi on the string (Fig. 5-22). Le t t b e the tension in the segment
PoPi, and Ti the tension in the segment PiP 2 . Let lo and h be the lengths
of these segments of the string, and let l 02 be the distance between P and
P 2 . The angles a, /3 between the two segments of string and the line PqP 2
are determined by the cosine law:
cos a =
02
+ ll - l\
21q1 2
cos /3 =
02
+ %- ll
21iIq2
(5-119)
236 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
Po
To
Fig. 5-22. A flexible string held at three points.
so that the position of the point Pi is independent of the force F x , provided
the string does not stretch. Since the bit of string at the point P x is in
equilibrium, the vector sum of the three forces Fi, t , and Ti acting on
the string at Pi must vanish, so that these forces form a closed triangle,
as indicated in Fig. 5-22. The tensions are then determine d in t erms of
the angle between the force F x and the direction of the line P0P2, by the
sine law:
sin (0 + 7) _ „ sin (7 - a) , ..
T ° = Fl sin(a + /3) ' Tl - Fl sin(«+0) (5 " 120)
Now suppose that the string stretches according to Hooke's law, so that
l = l' (i + fc To ), h = HO. + kri), (5-121)
where 1' , l[ are the unstretched lengths, and k is a constant [l//c would be
Young's modulus, Eq. (5-114), multiplied by the cross-sectional area of
the string]. The unknown quantities t , ti, l , and h can be eliminated
from Eqs. (5-119) by substitution from Eqs. (5-120) and (5-121). We
then have two rather complicated equations to be solved for the angles
a and /3. The solution must be carried out by numerical methods when
numerical values of 1' , l[, k, l 02 , F lt and 7 are given. When a and /3 are
found, r , Ti, Z , and h can be found from Eqs. (5-120) and (5-121). One
way of solving these equations by successive approximations is to assume
first that the string does not stretch, so that l = 1' , l 1 = l' lt and to calcu-
late a. and |3 from Eqs. (5-119), and t , Ti from Eqs. (5-120). Using these
values of t , ti, we then calculate l , h from Eqs. (5-121). The new
values of l , h can be used in Eqs. (5-119) to get better values for a, from
which better values of t , t x can be calculated. These can be used to get
still better values for l , h from Eqs. (5-121), and so on. As this process
is repeated, the successive calculated values of a, /3, To, ti, Iq, h will con-
verge toward the true values. If the string stretches only very little, the
5-9]
EQUILIBRIUM OF FLEXIBLE STRINGS AND CABLES
y
237
Fig. 5-23. A flexible string hanging under its own weight.
first few repetitions will be sufficient to give very close values. The method
suggested here is an example of a very general class of methods of solution
of physical problems by successive approximations. It is an example of
what are called relaxation methods of solving statics problems.
We next consider a string acted on by forces distributed continuously
along the length of the string. A point on the string will be specified by
its distance s from one end, measured along the string. Let f(s) be the
force per unit length at the point s, that is, the force on a small segment of
length ds is f ds. Then the total force acting on the length of string be-
tween the end s = and the point s is zero if the string is in equilibrium:
F + fids + T(s) = 0,
Jo
(5-122)
where F is the supporting force at the end s = 0, and t(s) is a vector
whose magnitude is the tension at the point s, oriented in the direction of
increasing s. By differentiating Eq. (5-122) with respect to s, we obtain
a differential equation for t(s) :
dr
ds
(5-123)
The simplest and most important application of Eq. (5-123) is to the
case of a string having a weight w per unit length. If the string is acted
on by no other forces except at the ends, it will hang in a vertical plane,
which we take to be the xy-plane, with the z-axis horizontal and the y-axis
vertical. Let d be the angle between the string and the x-axis (Fig. 5-23).
Then the horizontal and vertical components of Eq. (5-123) become :
^ (t sin 6) = w,
^(rcos*) = 0.
(5-124)
(5-125)
238 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
Equation (5-125) implies that
t cos 6 = C. (5-126)
The horizontal component of tension is constant, as it should be since the
external forces on the string are all vertical, except at the ends. By divid-
ing Eq. (5-124) by C, and using Eq. (5-126), we eliminate the tension:
-*- = c' (5_127)
If we represent the string by specifying the function y{x), we have the
relations
tan 6 = g = y', (5-128)
ds = [(dx) 2 + (dy) 2 ] 112 = dx(l + y' 2 ) 1 ' 2 , (5-129)
so that Eq. (5-127) becomes
^ = g ( l + (5-130)
This can be integrated, if w is constant :
sinh- 1 y' = ^ + a, (5-132)
where a is a constant. We solve for y' :
^' = != sinh (ir + *)- (5 " 133)
This can be integrated again, and we obtain
y= /3 + ^cosh0£ + «)- (5-134)
The curve represented by Eq. (5-134) is called a catenary, and is the form in
which a uniform string will hang if acted on by no force other than its own
weight, except at the ends. The constants C, /3, and a are to be chosen so
that y has the proper value at the endpoints, and so that the total length
of the string has the proper value. The total length is
1 = fds = f x Xl (1 + y' 2 ) 112 dx = J' 1 cosh 0£ + a) dx
JXq JXq
C
w
[sinh (VZL + a)- sinh (^ + a)] • (5-135)
5-10] EQUILIBRIUM OF SOLID BEAMS 239
5-10 Equilibrium of solid beams. A horizontal beam subject to vertical
forces is one of the simplest examples of a structure subject to shearing
forces and bending moments. To simplify the problem, we shall consider
only the case when the beam is under no compression or tension, and we
shall assume that the beam is so constructed and the forces so applied that
the beam bends in only one vertical plane, without any torsion (twisting)
about the axis of the beam. We find first the stresses within the beam
from a knowledge of the external forces, and then determine the distortion
of the beam due to these stresses.
Points along the beam will be located by a coordinate x measured hori-
zontally from the left end of the beam (Fig. 5-24). Let vertical forces
F\, . . . , F n act at the distances x\, . . . , x n from the left end. A force
will be taken as positive if it is directed upward. Let A A' be a plane
perpendicular to the beam at any distance x from the end. According
to Theorem I (5-107), of Section 5-6, the system of forces exerted across
the plane A A' by the material on the right against that on the left is
equivalent to a single force S through any point in the plane, and a couple
of torque N. (Note that in applying Theorem I, we are treating the
plane A A' as a rigid body, that is, we are assuming that the cross-sectional
plane AA' is not distorted by the forces acting on it.) In the case we are
considering there is no compression or tension and all forces are vertical, so
that S is directed vertically. We shall define the shearing force S as the
vertical force acting across A A' from right to left ; S will be taken as positive
when this force is directed upward, negative when it is downward.* By
Newton's third law, the force acting across AA' from left to right is —S.
Since we are assuming no torsion about the axis of the beam (z-axis), and
since all the forces are vertical, the torque N will be directed horizontally
and perpendicular to the beam. We shall define the bending moment N
4
^
-h' )n
Sflfi A ' \ Fs
Fig. 5-24. Forces acting on a beam.
* This sign convention for S is in agreement with sign conventions throughout
this book, where the upward direction is taken as positive. Sign conventions for
shearing force and bending moment are not uniform in physics and engineering
texts, and one must be careful in reading the literature to note what sign con-
vention is adopted by each author.
240 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
as the torque exerted from right to left across A A' about a horizontal axis
in the plane AA'; N will be taken as positive when it tends to rotate the
plane A A' in a counterclockwise direction. Since S is vertical, the torque
will be the same about any horizontal axis in the plane AA'.
The shearing force S and bending moment N can be determined by
applying the conditions of equilibrium [Eqs. (5-95) and (5-96)] to the
part of the beam to the left of the plane A A'. The total force and total
torque about a horizontal axis in the plane A A' are, if we neglect the
weight of the beam,
X) F t + S = 0, (5-136)
Xi<X
-No - X) (* - x *> F i + N = °> ( 5 - 137 )
X{<X
where the sums are taken over all forces acting to the left oi AA' and N
is the bending moment, if any, exerted by the left end of the beam against
its support. The torque iV will appear only if the beam is clamped or
otherwise fastened at its left end. The force exerted by any clamp or
other support at the end is to be included among the forces Fj. If the
beam has a weight w per unit length, this should be included in the
equilibrium equations :
J2 Fi - f X w dx + S = 0, (5-138)
x~ix J °
-N — J] (x — Xi)Fi + f X (x - x')w dx' + N = 0. (5-139)
xllx J °
The shearing force and bending moment at a distance x from the end are
therefore
S= - J2 Fi+ [ X w dx, (5-140)
x^tx J °
N = iV + Y* (* - x t )F t - f (x - x')w dx'. (5-141)
If there is any additional force distributed continuously along the beam,
this can be included in w as an additional weight per unit length. If the
beam is free at its ends, the shearing force and bending moment must be
zero at the ends. If we set aS = JV = at the right end of the beam,
equations (5-140) and (5-141) may be solved for two of the forces acting
on the beam when the others are known. If the beam is fastened or clamped
at either end, S and N may have any values there. Equations (5-140)
and (5-141) determine S and N everywhere along the beam when all
5-10]
EQUILIBRIUM OF SOLID BEAMS
241
Fig. 5-25. Distortion of a beam by shearing and bending, (a) Undistorted
beam, (b) Beam in shear, (c) Beam bent and in shear.
the forces are known, including the force and torque exerted through
the clamp, if any, on the left end. The shearing force and bending mo-
ment may be plotted as functions of a; whose slopes at any point are
obtained by differentiating Eqs. (5-140) and (5-141) :
dS
dx
= w,
(5-142)
dN - V F
lx~ - ±> Fi
Xi<X
f
JO
w dx' = — S.
(5-143)
The shearing force increases by — Fi from left to right across a point no-
where a force Fi acts.
Let us now consider the distortion produced by the shearing forces and
bending moments in a beam of uniform cross section throughout its
length. In Fig. 5-25(a) is shown an undistorted horizontal beam through
which are drawn a horizontal line 00' and a vertical plane AA'. In Fig.
5-25 (b) the beam is under a shearing strain, the effect of which is to slide
the various vertical planes relative to one another so that the line 00'
makes an angle 6 with the normal to the plane A A'. According to Eq.
(5-118), the angle is given in terms of the shearing force S and the shear
modulus n by:
S
e
nA
(5-144)
242 EIGID BODIES. ROTATION ABOUT AN AXIS. STATICS FcHAP. 5
0'
neutral
layer
Fig. 5-26. Strains in a bent beam.
where A is the cross-sectional area, and we have made the approximation
tan 6=6, since will be very small. In Fig. 5-25 (c), we show the further
effect of bending the beam. The plane AA' now makes an angle <p with
the vertical. It is assumed that the cross-sectional surface AA' remains
plane and retains its shape when the beam is under stress, although this
may not be strictly true near the points where forces are applied. In order
to determine <p, we consider two planes A A' and BB' initially vertical and
a small distance I apart. When the beam is bent, AA' and BB' will make
angles <p and <p + A<p with the vertical (Fig. 5-26). Due to the bending,
the fibers on the outside of the curved beam will be stretched and those on
the inside will be compressed. Somewhere within the beam will be a neu-
tral layer of unstretched fibers, and we shall agree to draw the line 00' so
that it lies in this neutral layer. A line between A A' and BB' parallel to
00' and a distance z above 00' will be compressed to a length I — Al,
where (see Fig. 5-26)
Al = z A<p. (5-145)
The compressive force dF exerted across an element of area dA a distance z
above the neutral layer 00' will be given by Eq. (5-114) in terms of Young's
modulus:
dF_
dA
= rf=K*
A<p
T
(5-146)
or, if we let I = ds, an infinitesimal element of length along the line 00',
(5-147)
4E. = Yz ^ •
dA ds
This equation is important in the design of beams, as it determines the
stress of compression or tension at any distance z from the neutral layer.
The total compressive force through the cross-sectional area A of the beam
5-10] EQUILIBRIUM OF SOLID BEAMS 243
will be
F = ffdF = Y^fJzdA. (5-148)
A A
Since we are assuming no net tension or compression of the beam, F = 0,
and
jjz dA = 0. (5-149)
This implies that the neutral layer contains the centroid of the area A of
the beam, and we may require that 00' be drawn through the centroid of
the cross-sectional area of the beam. The bending moment exerted by
the forces dF is
N = If'dF=Y%ff*dA
ds .
A
-.2 a d <P
= Yk'A^, (5-150)
where
k 2 = i- jjz 2 dA, (5-151)
A
and k is the radius of gyration of the cross-sectional area of the beam about
a horizontal axis through its centroid. The differential equation for <p is
therefore
d<p N
ds Yk 2 A
(5-152)
Let the upward deflection of the beam from a horizontal z-axis be y(x),
measured to the line 00' (Fig. 5-25). Then y(x) is to be determined by
solving the equation
^ = tan (6 + v ), (5-153)
when and <p have been determined from Eqs. (5-144) and (5-152). If
we assume that both and <p are very small angles, Eqs. (5-152) and
(5-153) become
S-7KT <-"*»
|-» + ». (5-155)
244 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
When there are no concentrated forces F, along the beam, we may differ-
entiate Eq. (5-155) and make use of Eqs. (5-154), (5-144), (5-142), and
(5-143) to obtain
d 2 y w . N ( 1 ft .
d^^^A + YWA' (5_156)
d^y_J_dSv_ w _ « im
«fa* _ nA dx 2 Yk 2 A K '
If bending can be neglected, as in a short, thick beam, Eq. (5-156) with
N = becomes a second-order differential equation to be solved for y(x).
For a longer beam, Eq. (5-157) must be used. These equations can also
be used when concentrated loads F* are present, by solving them for each
segment of the beam between the points where the forces Ft are applied,
and fitting the solutions together properly at these points. The solutions
on either side of a point Xi where a force F t - is applied must be chosen so
that y, <p, N are continuous across Xi, while S, dN/dx, dy/dx, d 3 y/dx 3
increase across the point X{ by an amount determined by Eqs. (5-140),
(5-143), (5-155), and (5-156). The solution of Eq. (5-156) will contain
two arbitrary constants, and that of Eq. (5-157), four, which are to be
determined by the conditions at the ends of the beam or segment of beam.
As an example, we consider a uniform beam of weight W, length L,
clamped in a horizontal position (i.e., so that <p = 0)* at its left end
(x = 0), and with a force Fi = —W exerted on its right end (x = L).
In this case, Eq. (5-157) becomes
t±-- W «-1«n
dx* ~ YWAL {b M)
The solution is
V=- 2iYk*AL + iCsx3 + * C2x2 + ClX + C °- (5_159)
To determine the constants Co, C\, C 2 , C3, we have at the left end of the
beam:
y = C = 0, (5-160)
%-<>>-'-& — *££• <«•»
where we have used Eqs. (5-155) and (5-144). We need two more con-
ditions, which may be determined in a variety of ways. The easiest way
* The condition <p = means that the plane A A' is vertical; that is, the beam
would be horizontal if there were no shearing strain.
5-11] EQUILIBRIUM OF FLUIDS 245
in this case is to apply Eq. (5-156) and its derivative at the left end of the
beam:
d 2 y r _ W W'L + WL ( , 1ft9 .
M> = ° 2 - ^AL TWA ' (5 ' 162)
d*y _ 1 dN S_ _ W' + W
dx* ~ U3 ~~ YWA dx ~ YWA ~ YWA ' v ° Q6>
where we have used Eq. (5-143). The deflection of the beam at any point
x is then
_ L 3
v YWA
Wx 2
AL 2
V 3I t 6L«/ t 2L2 V 1 3 L/l
-^i[-LV-2L) + -ry (5 " 164)
The deflection at x = L is
y = - Sx {W + * w ' ] -£a [w+ w ' ] - (5_i65)
The first term in each equation is the deflection due to bending, and the
second is that due to shear. The first term is proportional to L 3 , and in-
versely proportional to k 2 . The second term is proportional to L and in-
dependent of fc. Hence bending is more important for long, thin beams,
and shear is more important for short, thick beams. Our analysis here is
probably not very accurate for short, thick beams, since, as pointed out
above, some of our assumptions are not valid near points of support or
points where loads are applied (where "near" means relative to the cross-
sectional dimensions of the beam).
5-11 Equilibrium of fluids. A fluid is defined as a substance which will
support no shearing stress when in equilibrium. Liquids and gases fit this
definition, and even very viscous substances like pitch, or tar, or the mate-
rial in the interior of the earth, will eventually come to an equilibrium in
which shearing stresses are absent, if they are left undisturbed for a suffi-
ciently long time. The stress F/A across any small area A in a fluid in
equilibrium must be normal to A, and in practically all cases it will be a
compression rather than a tension.
We first prove that the stress F/A near any point in the fluid is inde-
pendent of the orientation of the surface A. Let any two directions be
given, and construct a small triangular prism with two equal faces A i =
A 2 perpendicular to the two given directions. The third face A 3 is to form
with A i and A 2 a cross section having the shape of an isosceles triangle
(Fig. 5-27). Let Fj, F 2 , F3 be the stress forces perpendicular to the faces
246 HIGID BODIES. BOTATION ABOUT AN AXIS. STATICS
[CHAP. 5
Fig. 5-27. Forces on a triangular prism in a fluid.
Ai, A 2 , A 3 . If the fluid in the prism is in equilibrium,
F x + F 2 + F 3 = 0.
(5-166)
The forces on the end faces of the prism need not be included here, since
they are perpendicular to F 1; F 2 , and F 3 , and must therefore separately
add to zero. It follows from Eq. (5-166), and from the way the prism has
been constructed, that F 1( F 2 , and F3 must form an isosceles triangle
(Fig. 5-27), and therefore that
Fi = F 2
(5-167)
Since the directions of Fx and F 2 are any two directions in the fluid, and
since Ai = A 2 , the stress F/A is the same in all directions. The stress in
a fluid is called the pressure p:
V
A x
El
A 2
(5-168)
Now suppose that in addition to the pressure the fluid is subject to an
external force f per unit volume of fluid, that is, any small volume dV in
the fluid is acted on by a force f dV. Such a force is called a body force;
f is the body force density. The most common example is the gravita-
tional force, for which
f = Pg, (5-169)
where g is the acceleration of gravity, and p is the density. In general,
the body force density may differ in magnitude and direction at different
points in the fluid. In the usual case, when the body force is given by
Eq. (5-169), g will be constant and f will be constant in direction; if p is
constant, f will also be constant in magnitude. Let us consider two nearby
points P u P 2 in the fluid, separated by a vector dr. We construct a cylinder
of length dt and cross-sectional area dA, whose end faces contain the points
Pi and P 2 . Then the total component of force in the direction of dt acting
5-11] EQUILIBRIUM OF FLUIDS 247
on the fluid in the cylinder, since the fluid is in equilibrium, will be
f 'dr dA + pi dA — p 2 dA = 0,
where pi and p 2 are the pressures at Pi and P 2 . The difference in pressure
between two points a distance dx apart is therefore
dp = p 2 — pi = f-dr. (5-170)
The total difference in pressure between two points in the fluid located by
vectors rx and r 2 will be
V2
-pi = Pt-dr, (5-171)
•'ri
where the line integral on the right is to be taken along some path lying
entirely within the fluid from rx to r 2 . Given the pressure pi at ri, Eq.
(5-171) allows us to compute the pressure at any other point r 2 which can
be joined to ri by a path lying within the fluid. The difference in pressure
between any two points depends only on the body force. Hence any
change in pressure at any point in a fluid in equilibrium must be accom-
panied by an equal change at all other points if the body force does not
change. This is Pascal's law.
According to the geometrical definition (3-107) of the gradient, Eq.
(5-170) implies that
f = Vp. (5-172)
The pressure gradient in a fluid in equilibrium must be equal to the body
force density. This result shows that the net force per unit volume due to
pressure is — Vp. The pressure p is a sort of potential energy per unit
volume in the sense that its negative gradient represents a force per unit
volume due to pressure. However, the integral of p dV over a volume
does not represent a potential energy except in very special cases. Equa-
tion (5-172) implies that the surfaces of constant pressure in the fluid are
everywhere perpendicular to the body force. According to Eqs. (3-187)
and (5-172), the force density f must satisfy the equation
V X f = 0. (5-173)
This is therefore a necessary condition on the body force in order for equi-
librium to be possible. It is also a sufficient condition for the possibility
of equilibrium. This follows from the discussion in Section 3-12, for if
Eq. (5-173) holds, then it is permissible to define a function p(r) by the
equation
P(r) = Pi + f t-dt, (5-174)
248 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
where Vi is the pressure at some fixed point rj, and the integral may be
evaluated along any path from r x to r within the fluid. If the pressure in
the fluid at every point r has the value p(r) given by (5-174), then Eq.
(5-172) will hold, and the body force f per unit volume will everywhere
be balanced by the pressure force — Vp per unit volume. Equation (5-174)
therefore defines an equilibrium pressure distribution for any body force
satisfying Eq. (5-173).
The problem of finding the pressure within a fluid in equilibrium, if the
body force density f (r) is given, is evidently mathematically identical with
the problem discussed in Section 3-12 of finding the potential energy for a
given force function F(r). We first check that V. X f is zero everywhere
within the fluid, in order to be sure that an equilibrium is possible. We
then take a point r x at which the pressure is known, and use Eq. (5-174) to
find the pressure at any other point, taking the integral along any conven-
ient path.
The total body force acting on a volume V of the fluid is
F 6 =
[[ft dV. (5-175)
The total force due to the pressure on the surface A of V is
F p = - jfnpdA, (5-176)
A
where n is the outward normal unit vector at any point on the surface.
These two must be equal and opposite, since the fluid is in equilibrium :
F p = -F 6 . (5-177)
Equation (5-176) gives the total force due to pressure on the surface of the
volume V, whether or not V is occupied by fluid. Hence we conclude from
Eq. (5-177) that a body immersed in a fluid in equilibrium is acted on by
a force F p due to pressure, equal and opposite to the body force F& which
would be exerted on the volume V if it were occupied by fluid in equi-
librium. This is Archimedes' principle. Combining Eqs. (5-172), (5-175),
(5-176), and (5-177), we have
ffnp dA = fffvp dV. (5-178)
A V
This equation resembles Gauss' divergence theorem [Eq. (3-115)], except
that the integrands are np and Vp instead of n-A and V-A. Gauss' the-
orem can, in fact, be proved in a very useful general form which allows us to
replace the factor n in a surface integral by V in the corresponding volume
5-11] EQUILIBRIUM OF FLUIDS 249
integral without any restrictions on the form of the integrand except that
it must be so written that the differentiation symbol V operates on the
entire integrand.* Given this result, we could start with Eqs. (5-175),
(5-176), and (5-177), and deduce Eq. (5-172) :
F 6 + F p = ffff dV - ffnp dA
V A
= f ff(i — Vp) dV = 0. (5-179)
v
Since this must hold for any volume V, Eq. (5-172) follows.
So far we have been considering only the pressure, i.e., the stress, in a
fluid. The strain produced by the pressure within a fluid is a change in
volume per unit mass of the fluid or, equivalently, a change in density.
If Hooke's law is satisfied, the change dV in a volume V produced by a
small change dp in pressure can be calculated from Eq. (5-116), if the
bulk modulus B is known :
d -~=~ d i- (5-180)
If the mass of fluid in the volume V is M , then the density is
M
P = y> (5-18D
and the change dp in density corresponding to an infinitesimal change dV
in volume is given by
*=-££, (5-182)
so that the change in density produced by a small pressure change dp is
* = $• (5-183)
After a finite change in pressure from p to p, the density will be
p = p exp(/^)- (5-184)
In any case, the density of a fluid is determined by its equation of state in
* For the proof of this theorem, see Phillips, Vector Analysis. New York:
John Wiley and Sons, 1933. (Chapter III, Section 34.)
250 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
terms of the pressure and temperature. The equation of state for a per-
fect gas is
pV = RT, (5-185)
where T is the absolute temperature, V is the volume per mole, and R is
the universal gas constant:
R = 8.314 X 10 7 erg-deg -1 C-mole -1 . (5-186)
By substitution from Eq. (5-181), we obtain the density in terms of pres-
sure and temperature :
P = f| > (5-187)
where M is the molecular weight.
Let us apply these results to the most common case, in which the body
force is the gravitational force on a fluid in a uniform vertical gravitational
field [Eq. (5-169)]. If we apply Eq. (5-173) to this case, we have
V X f = V X (/og) = 0. (5-188)
Since g is constant, the differentiation implied by the V symbol operates
only on p, and we can move the scalar p from one factor of the cross product
to the other to obtain :
(Vp) X g = 0, (5-189)
that is, the density gradient must be parallel to the gravitational field.
The density must be constant on any horizontal plane within the fluid.
Equation (5-189) may also be derived from Eq. (5-188) by writing
out explicitly the components of the vectors V X (pg) and (Vp) X g, and
verifying that they are the same.* According to Eq. (5-172), the pres-
sure is also constant in any horizontal plane within the fluid. Pressure
and density are therefore functions only of the vertical height z within the
fluid. From Eqs. (5-172) and (5-169) we obtain a differential equation
for pressure as a function of z:
f z = -pg. (5-190)
If the fluid is incompressible, and p is uniform, the solution is
V = Po — pgz, (5-191)
* Equation (5-189) holds also in a nonuniform gravitational field, since
T X g = 0, by Eq. (6-21).
PROBLEMS 251
where p is the pressure at z = 0. If the fluid is a perfect gas, either p or p
may be eliminated from Eq. (5-190) by means of Eq. (5-187). If we
eliminate the density, we have
g = _f£,. (5 _ 192)
As an example, if we assume that the atmosphere is uniform in temperature
and composition, we can solve Eq. (5-192) for the atmospheric pressure
as a function of altitude:
V~ RT 7
p = p exp i — -pf; z ) ■ (5-193)
Problems
1. (a) Prove that the total kinetic energy of the system of particles making up
a rigid body, as defined by Eq. (4-37), is correctly given by Eq. (5-16) when the
body rotates about a fixed axis, (b) Prove that the potential energy given by
Eq. (5-14) is the total work done against the external forces when the body is
rotated from d„ to 0, if N z is the sum of the torques about the axis of rotation due
to the external forces.
2. Prove, starting with the equation of motion (5-13) for rotation, that if N z
is a function of alone, then T + V is constant.
3. A wheel of mass M , radius of gyration k, spins smoothly on a fixed horizontal
axle of radius a which passes through a hole of slightly larger radius at the hub of
the wheel. The coefficient of friction between the bearing surfaces is /*. If the
wheel is initially spinning with angular velocity wo, find the time and the number
of turns that it takes to stop.
4. The balance wheel of a watch consists of a ring of mass M, radius a, with
spokes of negligible mass. The hairspring exerts a restoring torque N z = — kd.
Find the motion if the balance wheel is rotated through an angle do and released.
5. An airplane propeller of moment of inertia / is subject to a driving torque
N = No(l + acoswoO,
and to a frictional torque due to air resistance
N f = — b&.
Find its steady-state motion.
6. A motor armature weighing 2 kgm has a radius of gyration of 5 cm. Its
no-load speed is 1500 rpm. It is wound so that its torque is independent of its
speed. At full load, it draws a current of 2 amperes at 110 volts. Assume that
the electrical efficiency is 80%, and that the friction is proportional to the square
of the angular velocity. Find the time required for it to come up to a speed of
1200 rpm after being switched on without load.
7. Derive Eqs. (5-35) and (5-36).
252
RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
[CHAP. 5
8. Assume that a simple pendulum suffers a frictional torque — mbid due to
friction at the point of support, and a frictional force — 62V on the bob due to
air resistance, where v is the velocity of the bob. The bob has a mass m, and is
suspended by a string of length I. Find the time required for the amplitude to
damp to 1/e of its initial (small) value. How should m, I be chosen if it is desired
that the pendulum swing as long as possible? How should m, I be chosen if it
is desired that the pendulum swing through as many cycles as possible?
9. A compound pendulum is arranged to swing about either of two parallel
axes through two points 0, 0' located on a line through the center of mass. The
distances h, h! from 0, 0' to the center of mass, and the periods r, t' of small
amplitude vibrations about the axes through and 0' are measured. and 0'
are arranged so that each is approximately the center of oscillation relative to
the other. If r = r', find a formula for g in terms of measured quantities. If
t' = r(l + 8), where S <3C 1, find a correction to be added to your previous
formula so that it will be correct to terms of order S.
10. A baseball bat held horizontally at rest is struck at a point 0' by a ball
which delivers a horizontal impulse /' perpendicular to the bat. Let the bat be
initially parallel to the x-axis, and let the basbeall be traveling in the negative
direction parallel to the «/-axis. The center of mass G of the bat is initially at the
origin, and the point 0' is at a distance h' from G. Assuming that the bat is let
go just as the ball strikes it, and neglecting the effect of gravity, calculate and
sketch the motion x(t), y{t) of the center of mass, and also of the center of per-
cussion, during the first few moments after the blow, say until the bat has
rotated a quarter turn. Comment on the difference between the initial motion
of the center of mass and that of the center of percussion.
11. A circular disk of radius a lies in the zy-plane with its center at the origin.
The half of the disk above the z-axis has a density a per unit area, and the half
below the s-axis has a density 2<x. Find the center of mass G, and the moments of
inertia about the x-, y-, and 2-axes, and about parallel axes through G. Make as
much use of laborsaving theorems as possible.
12. (a) Work out a formula for the moments of inertia of a cone of mass m,
height h, and generating angle a, about its axis of symmetry, and about an axis
Fig. 5-28. Frustum of a cone.
PROBLEMS
253
45'
Fig. 5-29. How much thread can be wound on this spool?
through the apex perpendicular to the axis of symmetry. Find the center of mass
of the cone, (b) Use these results to determine the center of mass of the frustum
of a cone, shown in Fig. 5-28, and to calculate the moments of inertia about hori-
zontal axes through each base and through the center of mass. The mass of the
frustum is M .
13. How many yards of thread 0.03 inch in diameter can be wound on the
spool shown in Fig. 5-29?
14. Given that the volume of a cone is one-third the area of the base times the
height, locate by Pappus' theorem the centroid of a right triangle whose legs are
of lengths a and 6.
15. Prove that Pappus' second theorem holds even if the axis of revolution
intersects the surface, provided that we take as volume the difference in the
volumes generated by the two parts into which the surface is divided by the axis.
What is the corresponding generalization of the first theorem?
16. Find the center of mass of a wire bent into a semicircle of radius a. Find
the three radii of gyration about x-, y-, and 2-axes through the center of mass,
where z is perpendicular to the plane of the semicircle and x bisects the semicircle.
Use your ingenuity to reduce the number of calculations required to a minimum.
17. (a) Find a formula for the radius of gyration of a uniform rod of length I
about an axis through one end making an angle a with the rod. (b) Using this
result, find the moment of inertia of an equilateral triangular pyramid, con-
structed out of six uniform rods, about an axis through its centroid and one of
its vertices.
18. Find the radii of gyration of a plane lamina in the shape of an ellipse of
semimajor axis a, eccentricity e, about its major and minor axes, and about a
third axis through one focus perpendicular to the plane.
19. Forces 1 kgm-wt, 2 kgm-wt, 3 kgm-wt, and 4 kgm-wt act in sequence
clockwise along the four sides of a square 0.5 X 0.5 m 2 . The forces are directed
in a clockwise sense around the square. Find the equilibrant.
20. An iceboat has a flat sail in the shape of a right triangle with a vertical
leg of length o along the mast and a horizontal leg of length 6 along the boom.
The force on the sail acts at its centroid and is given by F = fc[n • (w — v)]n,
254 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS
1 4 lb
3 1b
[chap. 5
Fig. 5-30. A system of forces acting on a cube.
where n is a unit vector normal to the sail, w is the wind velocity, v is the velocity
of the boat, and k is a constant. The sail makes an angle a with the center line
of the boat. The angle a may have any value up to that for which F becomes
zero. The center line of the boat makes an angle /3 with the direction ( — w)
from which the wind is blowing. The runners are parallel to the center line.
The coefficient of friction along the runners is n, and there is a force N per-
pendicular to the runners sufficient to insure that v is parallel to the runners and
constant in direction. Find v as a function of a, /?, w and the mass m of the
boat. Find N and the point at which it acts. What value of a makes v a maxi-
mum if ji is very small?
21. (a) Reduce the system of forces acting on the cube shown in Fig. 5-30 to
an equivalent single force acting at the center of the cube, plus a couple com-
posed of two forces acting at two adjacent corners, (b) Reduce this system to
a system of two forces, and state where these forces act. (c) Reduce this system
to a single force plus a torque parallel to it.
22. (a) A cable is connected in a straight line between two fixed points. By
exerting a sidewise force W at the center of the cable, a considerably greater
force r can be applied to the support points at each end bf the cable. Find a
formula for t in terms of W, and the area A and Young's modulus Y of the
cable, assuming that the angle through which the cable is pulled is small, (b)
Show that this assumption is well satisfied if W = 100 lb, A = 3 in 2 , and
Y = 60,000 lb-in- 2 . Find t.
23. A cable is to be especially designed to hang vertically and to support a
load Fata distance I below the point of support. The cable is to be made of a
material having a Young's modulus F and a weight w per unit volume. Inas-
much as the length I of the cable is to be fairly great, it is desired to keep the
weight of the cable to a minimum by making the cross-sectional area A (z) of the
cable, at a height z above the lower end, just great enough to support the load
beneath it. The cable material can safely support a load just great enough to
stretch it 1%. Determine the function A(z) when the cable is supporting the
given load.
PROBLEMS
255
Fig. 5-31. A suspension bridge.
24. A cable 20 ft long is suspended between two points A and B, 15 ft apart.
The line AB makes an angle of 30° with the horizontal (B higher). A weight of
2000 lb is hung from a point C 8 ft from the end of the cable at A. (a) Find the
position of point C, and the tensions in the cable, if the cable does not stretch,
(b) If the cable is £ inch in diameter and has a Young's modulus of 5 X 10 5 lb-
in -2 , find the position of point C and the tensions, taking cable stretch into
account. Carry out two successive approximations, and estimate the accuracy
of your result.
25. (a) A cable of length I, weight w per unit length, is suspended from the
points x = ±aon the x-axis. The y-axis is vertical. By requiring that y = at
x = ±o, and that the total length of cable be I, show that a = in Eq. (5-134),
and set up equations to be solved for /3 and C. (b) Show that the same results
can be obtained for a and C by requiring that the cable be symmetrical about the
2/-axis, and that the forces at its ends balance the weight of the cable.
26. A bridge of weight w per unit length is to be hung from cables of negligible
weight, as shown in Fig. 5-31. It is desired to determine the shape of the suspen-
sion cables so that the vertical cables, which are equally spaced, will support
equal weights. Assume that the vertical cables are so closely spaced that we can
regard the weight w per unit length as continuously distributed along the
suspension cable. The problem then differs from that treated in the text, where
the string had a weight w per unit length s along the string, in that here there
is a weight w per unit horizontal distance x. Set up a differential equation for
the shape y (x) of the suspension cable, and solve for y(x) if the ends are at
the points y = 0, x = ±§Z), and if the maximum tension in the cable is to
be to-
27. A cable of length I, weight w per unit length, is suspended from points
x = =fca on the z-axis. The y-axts is vertical. A weight W is hung from the mid-
point of the cable. Set up the equations from which /3, a, and C are to be de-
termined.
28. A seesaw is made of a plank of wood of rectangular cross section 2X12 in 2
and 10 ft long, weighing 60 lb. Young's modulus is 1.5 X 10 6 lb-in -2 . The
plank is balanced across a narrow support at its center. Two children weighing
100 lb each sit one foot from the ends. Find the shape of the plank when it is
balanced in a stationary horizontal position. Neglect shear.
29. An empty pipe of inner radius a, outer radius b, is made of material with
Young's modulus Y, shear modulus n, density p. A horizontal section of length
L is clamped at both ends. Find the deflection at the center. Find the increase
in deflection when the pipe is filled with a fluid of density po.
256 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5
30. An I-beam has upper and lower flanges of width a, connected by a center
web of height 6. The web and flanges are of the same thickness c, assumed negligi-
ble with respect to a and b, and are made of a material with Young's modulus Y,
shear modulus n. The beam has a weight W , length L, and rests on supports at
each end. A load W rests on the midpoint of the beam. Find the deflection of
the beam at its midpoint. Separate the deflection into terms due to shear and
to bending, and into terms due to the beam weight W and the load W.
31. If the bulk modulus of water is B, and the atmospheric pressure at the sur-
face of the ocean is po, find the pressure as a function of depth in the ocean, taking
into account the compressibility of the water. Assume that B is constant. Look
up B for water, and estimate the error that would be made at a depth of 5 miles
if the compressibility were neglected.
32. Find the atmospheric pressure as a function of altitude on the assumption
that the temperature decreases with altitude, the decrease being proportional to
the altitude.
CHAPTER 6
GRAVITATION
6-1 Centers of gravity for extended bodies. You will recall that we
formulated the law of gravitation in Section 1-5. Any two particles of
masses wii and ra 2 , a distance r apart, attract each other with a force
whose magnitude is given by Eq. (1-11) :
F = ^ , (6-1)
where
G = 6.67 X 10 -8 dyne-cm 2 -gm -2 , (6-2)
as determined by measurements of the forces between large lead spheres,
carried out by means of a delicate torsion balance. Equation (6-1) can
be written in a vector form which gives both the direction and magnitude
of the attractive forces. Let ri and r 2 be the position vectors of the two
particles. Then the gravitational force on m 2 due to m x is
_ Gm 1 m 2 , r , ■ ( R o\
F i->2 = | fl _ r2 |3 ( f i - f 2)- (6-3)
The vector (r x — r 2 ) gives the force the correct direction, and its magni-
tude is divided out by the extra factor |r x — r 2 | in the denominator.
The law of gravitation as formulated in Eq. (6-3) is applicable only to
particles or to bodies whose dimensions are negligible compared with the
distance between them; otherwise the distance |r x — r 3 | is not precisely
denned, nor is it immediately clear at what points and in what directions
the forces act. For extended bodies, we must imagine each body divided
into pieces or elements, small compared with the distances between the
bodies, and compute the forces on each of the elements of one body due
to each of the elements of the other bodies.
Consider now an extended body of mass M and a particle of mass m
at a point P (Fig. 6-1). If the body of mass M is divided into small pieces
of masses m,-, each piece is attracted toward m by a force which we shall
call F,-. Now the system of forces F» can be resolved according to Theorem
I of Section 5-6 (5-107) into a single force through an arbitrary point,
plus a couple. Let this single force be F :
F = J2 F,, (6-4)
i
257
258 GRAVITATION [CHAP. 6
Fig. 6-1. Gravitational attraction between a particle and an extended body.
and let the arbitrary point be taken as the point P. Since none of the
forces Fj exerts any torque about P, the total torque about P is zero, and
the couple vanishes. The system of forces therefore has a resultant F
acting along a line through the mass m. The force acting on m is — F,
since Newton's third law applies to each of the forces F; in Eq. (6-4).
We locate on this line of action of F a point G a distance r from P such
that
|F| = ^- (6-5)
Then the system of gravitational forces between the body M and the par-
ticle m is equivalent to the single resultant forces F on M and — F on m
which would act if all the mass of the body M were concentrated at G.
The point G is called the center of gravity of the body M relative to the
point P; G is not, in general, at the center of mass of body M, nor even
on the Tine joining P with the center of mass. The parts of the body close
to P are attracted more strongly than those farther away, whereas in
finding the center of mass, all parts of the body are treated alike. Further-
more, the position of the point G will depend on the position of P. When
P is far away compared with the dimensions of the body, the acceleration
of gravity due to m will be nearly constant over the body and, in this
case, we showed in Section 5-6 that G will coincide with the center of
mass. Also, in the case of a uniform sphere or a spherically symmetrical
distribution of mass, we shah show in the next section that the center of
gravity always lies at the center of the sphere. The relative character of
the concept of center of gravity makes it of little use except in the case of
a sphere or of a body in a uniform gravitational held.
For two extended bodies, no unique centers of gravity can in general
be defined, even relative to each other, except in special cases, as when the
bodies are far apart, or when one of them is a sphere. The system of
6-2] GRAVITATIONAL FIELD AND GRAVITATIONAL POTENTIAL 259
gravitational forces on either body due to the other may or may not have
a resultant; if it does, the two resultants are equal and opposite and act
along the same line. However, even in this case, we cannot define defi-
nite centers of gravity (?i, G 2 for the two bodies relative to each other,
since Eq. (6-5) specifies only the distance GiG 2 -
The general problem of determining the gravitational forces between
bodies is usually best treated by means of the concepts of the field theory
of gravitation discussed in the next section.
6-2 Gravitational field and gravitational potential. The gravitational
force F m acting on a particle of mass m at a point r, due to other particles
m» at points r*, is the vector sum of the forces due to each of the other
particles acting separately:
F - - Z, | r . _ r |3 (6-6)
If, instead of point masses m*, we have mass continuously distributed in
space with a density p(r), the force on a point mass m at r is
-///
mG{ ?, ~ t (r0 dV. (6-7)
The integral may be taken over the region containing the mass whose
attraction we are computing, or over all space if we let p = outside
this region. Now the force F m is proportional to the mass m, and we
define the gravitational field intensity (or simply gravitational field) g(r),
at any point r in space, due to any distribution of mass, as the force per
unit mass which would be exerted on any small mass m at that point :
g(r) = — . (6-8)
m
where F m is the force that would be exerted on a point mass m at the
point r. We can write formulas for g(r) for point masses or continuously
distributed mass:
g(r) = E "f^ ^3 r) » (6-9)
The field g(r) has the dimensions of acceleration, and is in fact the
acceleration experienced by a particle at the point r, on which no forces
act other than the gravitational force.
260 GRAVITATION [CHAP. 6
The calculation of the gravitational field g(r) from Eq. (6-9) or (6-10)
is difficult except in a few simple cases, partly because the sum and in-
tegral call for the addition of a number of vectors. Since the gravitational
forces between pairs of particles are central forces, they are conservative,
as we showed in Section 3-12, and a potential energy can be denned for
a particle of mass m subject to gravitational forces. For two particles m
and mi, the potential energy is given by Eqs. (3-229) and (3-230) :
7 -< = ir?r7n (<W1)
The potential energy of a particle of mass m at point r due to a system of
particles m» is then
We define the gravitational potential g(r) at point r as the negative of the
potential energy per unit mass of a particle at point r: [This choice of
sign in g(r) is conventional in gravitational theory.]
8(r) = -^- (6-13)
For a system of particles,
* i *'
If p(r) represents a continuous distribution of mass, its gravitational poten-
tial is
w-fff&k'*'- <6 - 15)
Because it is a scalar point function, the potential g(r) is easier to work
with for many purposes than is the field g(r). In view of the relation
(3-185) between force and potential energy, g may easily be calculated,
when g is known, from the relation
g = VS. (6-16)
The inverse relation is
-f
S(r) = g-dr. (6-17)
The definition of g(r), like that of potential energy V(r), involves an arbi-
trary additive constant or, equivalently, an arbitrary point r, at which
g = 0. Usually r„ is taken at an infinite distance from all masses, as in
Eqs. (6-14) and (6-15).
6-2] GRAVITATIONAL FIELD AND GRAVITATIONAL POTENTIAL 261
Fig. 6-2. Method of computing potential of a spherical shell.
The concepts of gravitational field and gravitational potential are
mathematically identical to those of electric field intensity and electro-
static potential in electrostatics, except that the negative sign in Eq. (6-13)
is conventional in gravitational theory, and except that all masses are
positive and all gravitational forces are attractive, so that the force law
has the opposite sign from that in electrostatics. The subject of potential
theory is an extensive one, and we can give here only a very brief introduc-
tory treatment.
As an example of the use of the concept of potential, we calculate the
potential due to a thin homogeneous spherical shell of matter of mass M,
density a per unit area, and radius a:
M = 4ira 2 a. (6-18)
The potential at a point P is computed by integrating over a set of ring
elements as in Fig. 6-2. The potential of a ring of radius a sin 0, width
a dO, all of whose mass is at the same distance r from P, will be
, _ GaQira sin 9)a dO
and the total potential at P of the spherical shell is
9(P)
f G<r(2wa sin 0)a dd
Jo r
= MG J
2 Jo
sin dd
(r% + a 2 — 2ar o cos0) 112
= ^[(ro + o)- |ro-o|]- (6-19)
We have two cases, according to whether P is outside or inside the shell:
S(P) = ^ > r > a, g(P) = M£ , r < a. (6-20)
"o a
262 GRAVITATION [CHAP. 6
Thus outside the shell the potential is the same as for a point mass M at
the center of the shell. The gravitational field outside a spherical shell is
then the same as if all the mass of the shell were at its center. The same
statement then holds for the gravitational field outside any spherically
symmetrical distribution of mass, since the total field is the sum of the
fields due to the shells of which it is composed. This proves the statement
made in the previous section; a spherically symmetrical distribution of
mass attracts (and therefore is attracted by) any other mass outside it as
if all its mass were at its center. Inside a spherical shell, the potential is
constant, and it follows from Eq. (6-16) that the gravitational field is
there zero. Hence a point inside a spherically symmetric distribution of
mass at a distance r from the center is attracted as if the mass inside the
sphere of radius r were at the center; the mass outside this sphere exerts
no net force. These results would be somewhat more difficult to prove by
computing the gravitational forces directly, as the reader can readily
verify. Indeed, it took Newton twenty years! The calculation of the
force of attraction on the moon by the earth described in the last section
of Chapter 1 was made by Newton twenty years before he published his
law of gravitation. It is likely that he waited until he could prove an
assumption implicit in that calculation, namely, that the earth attracts any
body outside it as if all the mass of the earth were concentrated at its center.
6-3 Gravitational field equations. It is of interest to find differential
equations satisfied by the functions g(r) and g(r). From Eq. (6-16) it
follows that
V X g = 0. (6-21)
When written out in any coordinate system, this vector equation becomes
a set of three partial differential equations connecting the components of
the gravitational field. In rectangular coordinates,
a^-^ =0 ' "to ~ to ~ °' to _ dy - °- (6 ~ 22)
These equations alone do not determine the gravitational field, for they
are satisfied by every gravitational field. To determine the gravitational
field, we need an equation connecting g with the distribution of matter.
Let us study the gravitational field g due to a point mass m. Consider
any volume V containing the mass m, and let n be the unit vector normal
at each point to the surface S that bounds V (Fig. 6-3). Let us compute
the surface integral
I = ffn-gdS. (6-23)
6-3] GRAVITATIONAL FIELD EQUATIONS 263
Fig. 6-3. A mass m enclosed in a volume V.
The physical or geometric meaning of this integral can be seen if we in-
troduce the concept of hues of force, drawn everywhere in the direction
of g, and in such a manner that the number of lines per square centimeter
at any point is equal to the gravitational field intensity. Then I is the
number of lines passing out through the surface 5, and is called the flux
of g through 8, The element of solid angle dti subtended at the position
of m fay an clement of surface dS is defined as the area swept out on a
sphere of unit radius by a, radius from m which sweeps over the surface
element dS. This area is
,„ dS cos B ,„ _,,
dU = r2 ■ (6-24)
Prom Fig. 6-3, we have the relation
mG cos 9
n-g= ^— ■ (6-25)
When use is made of these two relations, the integral I [Eq. (0-23)] be-
comes
/ = f f—mGdtt = — 4nmG. (6-26)
The integral 7 is independent of the position of m within the surface S.
This result is analogous to the corresponding result in electrostatics that
there are 4ir lines of force coming from every unit charge. Since the
gravitational field of a number of masses is the sum of their individual
fields, we have, for a surface 8 surrounding a set of masses rrii :
I = f fn-gdS = — J^ 4irmiG. (6-27)
8 i
For a continuous distribution of mass within S, this equation becomes
ffn-gdS^ - f f fiirGp dV. (i 5-28 !
264 GRAVITATION [CHAP. 6
We now apply Gauss' divergence theorem [Eq. (3-115)] to the left side of
this equation:
Jfn-gdS = fffv-gdV. (6-29)
s v
Subtracting Eq. (6-28) from Eq. (6-29), we arrive at the result
///(V-g + 4arGp) dV = 0. (6-30)
v
Now Eq. (6-30) must hold for any volume V, and this can only be true if
the integrand vanishes:
V • g = — 4irGp. (6-31)
This equation in cartesian coordinates has the form
?gi + d Jv + d Jl = -4arGp(x, y, z). (6-32)
dx dy dz
When p(x, y, z) is given, the set of equations (6-22) and (6-32) can be
shown to determine the gravitational field (g x , g v , g z ) uniquely, if we add
the boundary condition that g -» as |r| -> oo. Substituting from Eq.
(6-16), we get an equation satisfied by the potential:
V 2 g = -4tt(?p, (6-33)
or
i2,
^8 + ^S + ^8 = _ 47r (?p. (6-34)
dx 2 + dy2 ^ dz 2 p
This single equation determines <3(x, y, z) uniquely if we add the condition
that g -> as |r| -> oo. This result we quote from potential theory with-
out proof. The solution of Eq. (6-33) is, in fact, Eq. (6-15). It is often
easier to solve the partial differential equation (6-34) directly than to com-
pute the integral in Eq. (6-15). Equations (6-33), (6-16), and (6-8) to-
gether constitute a complete summary of Newton's theory of gravitation,
as likewise do Eqs. (6-31), (6-21), and (6-8) ; that is, all the results of the
theory can be derived from either of these sets of equations.
Equation (6-33) is called Poisson's equation. Equations of this form
turn up frequently in physical theories. For example, the electrostatic
potential satisfies an equation of the same form, where p is the electric
charge density. If p = 0, Eq. (6-33) takes the form
V 2 9 = 0. (6-35)
This is called Laplace's equation. An extensive mathematical theory of
PROBLEMS 265
Eqs. (6-33) and (6-35) has been developed.* A discussion of potential
theory is, however, outside the scope of this text.
Problems
1. (a) Given Newton's laws of motion, and Kepler's first two laws of planetary
motion (Section 3-15), show that the force acting on a planet is directed toward
the sun and is inversely proportional to the square of the distance from the sun.
(b) Use Kepler's third law to show that the forces on the planets are propor-
tional to their masses, (c) If this suggests to you a universal law of attraction
between any two masses, use Newton's third law to show that the force must be
proportional to both masses.
2. (a) Find the gravitational field and gravitational potential at any point z
on the symmetry axis of a uniform solid hemisphere of radius a, mass M . The
center of the hemisphere is at z = 0. (b) Locate the center of gravity of the
hemisphere relative to a point outside it on the z-axis, and show that as
z -> ±oo, the center of gravity approaches the center of mass.
3. Assuming that the earth is a sphere of uniform density, with radius a,
mass M , calculate the gravitational field intensity and the gravitational potential
at all points inside and outside the earth, taking 9 = at an infinite distance.
4. Assuming that the interior of the earth can be treated as an incompressible
fluid in equilibrium, (a) calculate the pressure within the earth as a function of
distance from the center, (b) Using appropriate values for the earth's mass and
radius, calculate the pressure in tons per square inch at the center.
5. Show that if the sun were surrounded by a spherical cloud of dust of uniform
density p, the gravitational field within the dust cloud would be
(MG . 4x _\r
e= -\-W + Y pGr )r'
where M is the mass of the sun, and r is a vector from the sun to any point in the
dust cloud.
6. Assume that the density of a star is a function only of the radius r measured
from the center of the star, and is given by f
Ma, 2
2nr(r2+a 2 ) 2
where M is the mass of the star, and a is a constant which determines the size of
the star. Find the gravitational field intensity and the gravitational potential as
functions of r.
* O. D. Kellogg, Foundations of Potential Theory. Berlin: J. Springer, 1929.
t The expression for p is chosen to make the problem easy to solve, not because
it has more than a remote resemblance to the density variations within any actual
star.
266 GRAVITATION [CHAP. 6
7. Set up the equations to be solved for the pressure as a function of radius
in a spherically symmetric mass M of gas, assuming that the gas obeys the perfect
gas laws and that the temperature is known as a function of radius.
8. (a) Assume that ordinary cold matter collapses, under a pressure greater
than a certain critical pressure po, to a state of very high density pi. A planet of
mass M is constructed of matter of mean density po in its normal state. Assuming
uniform density and conditions of fluid equilibrium, at what mass M o and radius
ro will the pressure at the center reach the critical value po? (b) li M > Mo, the
planet will have a very dense core of density pi surrounded by a crust of density
po. Calculate the resulting pressure distribution within the planet in terms of the
radius r\ of the core and the radius r 2 of the planet. Show that if M is somewhat
larger than Mo, then the radius ri of the planet is less than ro. (The planet Jupiter
is said to have a mass very nearly equal to the critical mass Mo, so that if it were
heavier it might be smaller.)
9. Find the pressure and temperature as functions of radius for the star of
Problem 6 if the star is composed of a perfect gas of atomic weight A.
10. Find the density and gravitational field intensity as a function of radius
inside a small spherically symmetric planet, to order (1/B 2 ), assuming that the
bulk modulus B is constant. The mass is M and the radius is a. [Hint: Calculate
g(r) assuming uniform density; then find the resulting pressure p(r), and the
density p(r) to order (1/B). Recalculate g(r) using the new p(r), and proceed
by successive approximations to terms of order (1/B 2 ).]
11. Consider a spherical mountain of radius a, mass M, floating in equilibrium
in the earth, and whose density is half that of the earth. Assume that a is much
less than the earth's radius, so that the earth's surface can be regarded as fiat in
the neighborhood of the mountain. If the mountain were not present, the gravi-
tational field intensity near the earth's surface would be go- (a) Find the differ-
ence between go and the actual value of g at the top of the mountain, (b) If the
top of the mountain is eroded flat, level with the surrounding surface of the earth,
and if this occurs in a short time compared with the time required for the moun-
tain to float in equilibrium again, find the difference between go and the actual
value of g at the earth's surface at the center of the eroded mountain.
12. (a) Find the gravitational potential and the field intensity due to a thin
rod of length I and mass M at a point a distance r from the center of the rod in a
direction making an angle 8 with the rod. Assume that ry> I, and carry the
calculations only to second order in l/r. (b) Locate the center of gravity of the
rod relative to the specified point.
13. (a) Calculate the gravitational potential of a uniform circular ring of
matter of radius a, mass M, at a distance r from the center of the ring in a direc-
tion making an angle with the axis of the ring. Assume that r y> a, and cal-
culate the potential only to second order in a/r. (b) Calculate to the same ap-
proximation the components of the gravitational field of the ring at the specified
point.
14. A small body with cylindrical symmetry has a density p(r, 0) in spherical
coordinates, which vanishes for r > a. The origin r = lies at the center of
mass. Approximate the gravitational potential at a point r, far from the body
PROBLEMS 267
(r ^> a), by expanding in a power series in (a/r), and show that it has the form
9(r,«) =^ + ^P2(cos0) + ^P 3 (cos0) + ---,
where P2 (cos 6), P3 (cos 0) are quadratic and cubic polynomials in cos that
do not depend on the body, and Q, E are constants which depend on the mass
distribution. Find expressions for P2, P3, Q, and E, and show that Q is of the
order of magnitude Ma 2 , and E of the order Ma 3 . It is conventional to normalize
Pi so that the constant term is —\, and Pz so that the linear term is — § cos 0.
The parameters Q, E are then called the quadrupole moment and the octopole
moment of the body. The polynomials P2, P3, • ■ • are the Legendre polynomials.
15. The earth has approximately the shape of an oblate ellipsoid of revolution
whose polar diameter 2a(l — tj) is slightly shorter than its equatorial diameter
2a. (ij = 0.0034.) To determine to first order in rj, the effect of the earth's
oblateness on its gravitational field, we may replace the ellipsoidal earth by a
sphere of radius R so chosen as to have the same volume. The gravitational
field of the earth is then the field of a uniform sphere of radius R with the mass
of the earth, plus the field of a surface distribution of mass (positive or negative),
representing the mass per unit area which would be added or subtracted to form
the actual ellipsoid.
(a) Show that the required surface density is, to first order in rj,
a = ^ijap(l — 3 cos 2 0),
where is the colatitude, and p is the volume density of the earth (assumed
uniform). Since the total mass thus added to the surface is zero, its gravitational
field will represent the effect of the oblate shape of the earth.
(b) Show that the resulting correction to the gravitational potential at a very
great distance r^> a from the earth is, to order (a 3 //- 3 ),
ss ~ \ v ^r~ (1 ~ 3 cos2 6) ' {r y> a) -
16. (a) Use Gauss' theorem (6-26) to determine the gravitational field inside
and outside a spherical shell of radius a, mass M, uniform density, (b) Cal-
culate the resulting gravitational potential.
17. (a) Find the gravitational field at a distance x from an infinite plane
sheet of density a per unit area, (b) Compare this result with the field just
outside a spherical shell of the same surface density. What part of the field
comes from the immediately adjacent matter and what part from more distant
matter?
18. Show that the gravitational field equations (6-21), (6-31), and (6-33)
are satisfied by the field intensity and potential which you calculated in Prob-
lem 3.
*19. (a) Show that SQ found in Problem 15(b) satisfies Laplace's equation
(6-35). This, together with the fact that 59 has the same angular dependence
268 GRAVITATION [CHAP. 6
as the mass density which produces it, suggests that the formula given for 5Q
may actually be valid everywhere outside the earth, (b) To show this, consider
Poisson's equation (6-33) with p — /(r)(l — 3cos 2 0). Show that a solution
9 = *(r)(l — 3 cos 2 0) will satisfy Eq. (6-33) with this form of p, provided
d 2 h , 2 dh &h , _,
(c) Show that h = r -3 satisfies this equation in the region where / = 0. Can
you complete the proof that the formula for 5g found in Problem 15(b) is in
fact valid everywhere outside the earth?
CHAPTER 7
MOVING COORDINATE SYSTEMS
7-1 Moving origin of coordinates. Let a point in space be located by
vectors r, r* with respect to two origins of coordinates 0, 0*, and let 0* be
located by a vector h with respect to
(Fig. 7-1). Then the relation be-
tween the coordinates r and r* is
given by
r = r* + h, (7-1)
r* = r
(7-2)
In terms of rectangular coordinates,
with axes x*, y*, z* parallel to axes
x, y, z, respectively, these equations
can be written:
Fig. 7-1. Change of origin of coordi-
nates.
x = x* + h x
x* = x
y = y* + h v ,
y* = y — h y ,
2 = z* + h z ; (7-3)
z* = z- h z . (7-4)
Now if the origin 0* is moving with respect to the origin 0, which we
regard as fixed, the relation between the velocities relative to the two sys-
tems is obtained by differentiating Eq. (7-1) :
_ dx _ drj* cfli
dt dt dt
= v* + Vft,
(7-5)
where v and v* are the velocities of the moving point relative to and 0*,
and Vh is the velocity of 0* relative to 0. We are supposing that the
axes x*, y*, z* remain parallel to x, y, z. This is called a translation of the
starred coordinate system with respect to the unstarred system. Written
out in cartesian components, Eq. (7-5) becomes the time derivative of
Eq. (7-3). The relation between relative accelerations is
d 2 T
2_*
d't
= a* + a A .
d\
df2
(7-6)
Again these equations can easily be written out in terms of their rectangu-
lar components.
269
270 MOVING COORDINATE SYSTEMS [CHAP. 7
Newton's equations of motion hold in the fixed coordinate system, so
that we have, for a particle of mass m subject to a force F:
m § = F " ™
Using Eq. (7-6), we can write this equation in the starred coordinate sys-
tem:
m ^p- + ma h = F. (7-8)
If 0* is moving at constant velocity relative to 0, then &h = 0, and we
have
m *±- = F. (7-9)
Thus Newton's equations of motion, if they hold in any coordinate system,
hold also in any other coordinate system moving with uniform velocity
relative to the first. This is the Newtonian principle of relativity. It
implies that, so far as mechanics is concerned, we cannot specify any
unique fixed coordinate system or frame of reference to which Newton's
laws are supposed to refer; if we specify one such system, any other system
moving with constant velocity relative to it will do as well. This property
of Eq. (7-7) is sometimes expressed by saying that Newton's equations of
motion remain invariant in form, or that they are covariant, with respect to
uniform translations of the coordinates. The concept of frame of reference
is not quite the same as that of a coordinate system, in that if we make a
change of coordinates that does not involve the time, we do not regard this
as a change of frame of reference. A frame of reference includes all coordi-
nate systems at rest with respect to any particular one. The principle
of relativity proposed by Einstein asserts that the relativity principle is not
restricted to mechanics, but holds for all physical phenomena. The special
theory of relativity is the result of the application of this principle to all
types of phenomena, particularly electromagnetic phenomena. It turns
out that this can only be done by modifying Newton's equations of motion
slightly and, in fact, even Eqs. (7-5) and (7-6) require modification, f
For any motion of 0*, we can write Eq. (7-8) in the form
= F — raa A . (7-10)
dP
This equation has the same form as the equation of motion (7-7) in a fixed
coordinate system, except that in place of the force F, we have F — m&h-
t P. G. Bergmann, Introduction to the Theory of Relativity. New York: Prentice-
Hall, 1946. (Part 1.)
7-2] ROTATING COORDINATE SYSTEMS 271
The term —m&h we may call a fictitious force. We can treat the motion
of a mass m relative to a moving coordinate system using Newton's equa-
tions of motion if we add this fictitious force to the actual force which acts.
From the point of view of classical mechanics, it is not a force at all, but
part of the mass times acceleration transposed to the other side of the
equation. The essential distinction is that the real forces F acting on m
depend on the positions and motions of other bodies, whereas the fictitious
force depends on the acceleration of the starred coordinate system with
respect to the fixed coordinate system. In the general theory of relativity,
terms like -maj are regarded as legitimate forces in the starred coordi-
nate system, on the same footing with the force F, so that in all coordinate
systems the same law of motion holds. This, of course, can only be done
if it can be shown how to deduce the force —ma^ from the positions and
motions of other bodies. The program is not so simple as it may seem
from this brief outline, and modifications in the laws of motion are required
to carry it through, f
7-2 Rotating coordinate systems. We now consider coordinate systems
x, y, z and x*, y*, z* whose axes are rotated relative to one another as in
Fig. 7-2, where, for the present, the origins of the two sets of axes coin-
cide. Introducing unit vectors i, j, k associated with axes x, y, z, and
unit vectors i*, j*, k* associated with axes x*, y*, z*, we can express the
position vector r in terms of its components along either set of axes:
r = xx + y] + zk, (7-11)
r = x *i* + y*j* + z*k*. (7-12)
Note that since the origins now coincide, a point is represented by the
same vector r in both systems; only the components of r are different
along the different axes. The relations between the coordinate systems
Fig. 7-2. Rotation of coordinate axes.
t Bergmann, op. cit. (Part 2.)
272 MOVING COORDINATE SYSTEMS [CHAP. 7
can be obtained by taking the dot product of either the starred or the
unstarred unit vectors with Eqs. (7-11) and (7-12). For example, if we
compute i-r, j-r, k-r, from Eqs. (7-11) and (7-12) and equate the results,
we obtain __ , _ „
x = s*(i*.i) + 2/*(j*-i) + 2*(k*-i),
y = s*(i*-j) + y*(j*-J) + «*(k*'J), (7-13)
z = x*(i*-k) + »*0*-k) + z*(k*-k).
The dot products (i*-i), etc., are the cosines of the angles between the
corresponding axes. Similar formulas for x*, y*, z* in terms of x, y, z can
easily be obtained by the same process. These formulas are rather compli-
cated and unwieldy, and we shall fortunately be able to avoid using them
in most cases. Equations (7-11), (7-12), and (7-13) do not depend on the
fact that the vector r is drawn from the origin. Analogous formulas apply
in terms of the components of any vector A along the two sets of axes.
The time derivative of any vector A was denned by Eq. (3-52) :
dA =lim A(t + M)-A(t) . (y _ 14)
at aj->o At
In attempting to apply this definition in the present case, we encounter
a difficulty if the coordinate systems are rotating with respect to each
other. A vector which is constant in one coordinate system is not con-
stant in the other, but rotates. The definition requires us to subtract
A(0 from A(t + At). During the time At, coordinate system x*, y*, z* has
rotated relative to x, y, z, so that at time t + At, the two systems will not
agree as to which vector is (or was) A(0, i.e., which vector is in the same
position that A was in at time t. The result is that the time derivative of
a given vector will be different in the two coordinate systems. Let us use
d/dt to denote the time derivative with respect to the unstarred coordinate
system, which we regard as fixed, and d*/dt to denote the time derivative
with respect to the rotating starred coordinate system. We make this
distinction with regard to vectors only; there is no ambiguity with regard
to numerical quantities, and we denote their time derivatives by d/dt, or
by a dot, which will have the same meaning in all coordinate systems.
Let the vector A be given by
A = A x i + A y i + A*k, (7-15)
A = A%i* + A*j* + A%*. (7-16)
The unstarred time derivative of A may be obtained by differentiating
Eq. (7-15), regarding i, j, k as constant vectors in the fixed system:
^ = A x i + A v j + AX (7-17)
7-2]
ROTATING COORDINATE SYSTEMS
273
Similarly, the starred derivative of A is given in terms of its starred com-
ponents by
£p = i*i* + A%* -r Atk*.
(7-18)
We may regard Eqs. (7-17) and (7-18) as the definitions of unstarred and
starred time derivatives of a vector. We can also obtain a formula for
d/dt in starred components by taking the unstarred derivative of Eq.
(7-16), remembering that the unit vectors i*, j*, k* are moving relative
to the unstarred system, and have time derivatives:
dA
dt
A%i*
+ A*j* + Atk* + At^- + A
*dj*_
y dt
+ A ° dt
(7-19)
A similar formula could be obtained for d*A/dt in terms of its unstarred
components.
Let us now suppose that the starred coordinate system is rotating about
some axis OQ through the origin, with an angular velocity w (Fig. 7-3).
We define the vector angular velocity u as a vector of magnitude w directed
along the axis OQ in the direction of advance of a right-hand screw rotat-
ing with the starred system. Consider a vector B at rest in the starred
system. Its starred derivative is zero, and we now show that its unstarred
derivative is
dB
dt
= w x B.
(7-20)
In order to subtract B(t) from B(t + At), we draw these vectors with
their tails together, and it will be convenient to place them with their
tails on the axis of rotation. (The time derivative depends only on the com-
ponents of B along the axes, and not on the position of B in space.) We
Fig. 7-3. Time derivative of a rotating vector.
274 MOVING COORDINATE SYSTEMS [CHAP. 7
first verify from Fig. 7-3 that the direction of dB/dt is given correctly by
Eq. (7-20), recalling the definition [Eq. (3-24) and Fig. 3-11] of the cross
product. The magnitude of dB/dt as given by Eq. (7-20) is
dB
dt
\a X B| = uB sin 6. (7-21)
This is the correct formula, since it can be seen from Fig. 7-3 that, when
At is small,
|AB| = (.Bsinfl)(wAO.
When Eq. (7-20) is applied to the unit vectors i*, j*, k*, Eq. (7-19) be-
comes, if we make use of Eqs. (7-18) and (7-16) :
dA d*A
^ = ~ + A* x (a x i*) + A* y (a x j*) + A* z (a x k*)
d*A
= ~ + a X A. (7-22)
dt
This is the fundamental relationship between time derivatives for rotating
coordinate systems. It may be remembered by noting that the time de-
rivative of any vector in the unstarred coordinate system is its derivative
in the starred system plus the unstarred derivative it would have if it
were at rest in the starred system. Equation (7-22) applies even when
the angular velocity vector a is changing in magnitude and direction with
time. Taking the derivative of right and left sides of Eq. (7-22), and
applying Eq. (7-22) again to A and d*A/dt, we have for the second time
derivative of any vector A :
dA . da .
d*A = d (d*A\
dt 2 ~ dt\dt )
d*A , (d*A . .\ , dw w .
+ wX nr + » x \-dT + wxA ) + -di xA
_rf* 2 A
dt 2
d* 2 A , „ d*A , , ., . dw . ,_ „„.
= _ r + 2wX __ +wX(wxA )+^xA. (7-23)
In view of Eq. (3-29), the starred and unstarred derivatives of any vector
parallel to the axis of rotation are the same, according to Eq. (7-22). In
particular,
da _ d*a
~dl ~ IT'
7-2] ROTATING COORDINATE SYSTEMS 275
It is to be noted that the vector w on both sides of this equation is the
angular velocity of the starred system relative to the unstarred system,
although its time derivative is calculated with respect to the unstarred
system on the left side, and with respect to the starred system on the
right. The angular velocity of the unstarred system relative to the starred
system will be — to.
We now show that the relations derived above for a rotating coordinate
system are perfectly general, in that they apply to any motion of the
starred axes relative to the unstarred axes. Let the unstarred rates of
change of the starred unit vectors be given in terms of components along
the starred axes by
di*
-£ = ani* + a 12 j* + ai 3 k*,
d\*
-jj- = a 21 i* + a 22 j* + a 23 k*, (7-24)
dk*
-jf = a 3 ii* + a 32 j* + a 33 k*.
By differentiating the equation
i*-i* = 1, (7-25)
we obtain
f -i* = 0. (7-26)
From this and the corresponding equations for j* and k*, we have
an = a 22 = a 33 = 0. (7-27)
By differentiating the equation
i*-k* = 0, (7-28)
we obtain
«K* = — 1 • ■
dt dt
(7-29)
From this and the other two analogous equations, we have
«31 = — ai3, 0, 12 = — 021, 023 = —a32- (7-30)
Let a vector to be defined in terms of its starred components by:
w* = a 23 , 03% = o 31 , co^ = a 12 . (7-31)
di*
dt
=
CO
X
i*,
dj*
dt
=
CO
X
i*,
dk*
dt
=
w
X
k*.
276 MOVING COORDINATE SYSTEMS [CHAP. 7
Equations (7-24) can now be rewritten, with the help of Eqs. (7-27),
(7-30), and (7-31), in the form
(7-32)
According to Eq. (7-20), these time derivatives of i*, j*, k* are just those
to be expected if the starred unit vectors are rotating with an angular
velocity a. Thus no matter how the starred coordinate axes may be
moving, we can define at any instant an angular velocity vector «, given
by Eq. (7-31), such that the time derivatives of any vector relative to the
starred and unstarred coordinate systems are related by Eqs. (7-22) and
(7-23).
Let us now suppose that the starred coordinate system is moving so that
its origin 0* remains fixed at the origin of the fixed coordinate system.
Then any point in space is located by the same position vector r in both
coordinate systems [Eqs. (7-11) and (7-12)]. By applying Eqs. (7-22)
and (7-23) to the position vector r, we obtain formulas for the relation
between velocities and accelerations in the two coordinate systems:
dx d*T ,„ .
dt == W + MXI > (7 " 33)
d 2 i d* 2 t , , . , „ d*r , du ,„ „,.
dii = -dW + » x (» x ^+ 2 « x W+dt XI - (7 " 34)
Formula (7-34) is called Coriolis' theorem. The first term on the right is
the acceleration relative to the starred system. The second term is called
the centripetal acceleration of a point in rotation about an axis (centripetal
means "toward the center"). Using the notation in Fig. 7-4, we readily
verify that to X (« X r) points directly toward and perpendicular to the
axis of rotation, and that its magnitude is
\w X (« X r)| = u 2 r sine
v 2
(7-35)
rsin 9
where v = ur sin 6 is the speed of circular motion and (r sin 6) is the dis-
tance from the axis. The third term is present only when the point r is
moving in the starred system, and is called the coriolis acceleration. The
7-2] ROTATING COORDINATE SYSTEMS 277
Fig. 7-4. Centripetal acceleration.
last term vanishes for a constant angular velocity of rotation about a
fixed axis.
If we suppose that Newton's law of motion (7-7) holds in the unstarred
coordinate system, we shall have in the starred system:
m^~ + ma X (w X r) + 2wuo X ^~ + m^ Xi = F. (7-36)
Transposing the second, third, and fourth terms to the right side, we ob-
tain an equation of motion similar in form to Newton's equation of motion :
to -—- = F — mwx(wxr)— 2mo» X -3- m -^ X r. (7-37)
The second term on the right is called the centrifugal force (centrifugal
means "away from the center"); the third term is called the coriolis force.
The last term has no special name, and appears only for the case of non-
uniform rotation. If we introduce the fictitious centrifugal and coriolis
forces, the laws of motion relative to a rotating coordinate system are the
same as for fixed coordinates. A great deal of confusion has arisen regard-
ing the term "centrifugal force." This force is not a real force, at least
in classical mechanics, and is not present if we refer to a fixed coordinate
system in space. We can, however, treat a rotating coordinate system as
if it were fixed by introducing the centrifugal and coriolis forces. Thus a
particle moving in a circle has no centrifugal force acting on it, but only a
force toward the center which produces its centripetal acceleration. How-
ever, if we consider a coordinate system rotating with the particle, in this
system the particle is at rest, and the force toward the center is balanced
by the centrifugal force. It is very often useful to adopt a rotating coordi-
nate system. In studying the action of a cream separator, for example, it
is far more convenient to choose a coordinate system in which the liquid
278 MOVING COORDINATE SYSTEMS [CHAP. 7
is at rest, and use the laws of diffusion to study the diffusion of cream
toward the axis under the action of the centrifugal force field, than to try
to study the motion from the point of view of a fixed observer watching
the whirling liquid.
We can treat coordinate systems in simultaneous translation and rota-
tion relative to each other by using Eq. (7-1) to represent the relation be-
tween the coordinate vectors r and r* relative to origins 0, 0* not neces-
sarily coincident. In the derivation of Eqs. (7-32), no assumption was
made about the origin of the starred coordinates, and therefore Eqs. (7-22)
and (7-23) may still be used to express the time derivatives of any vector
with respect to the unstarred coordinate system in terms of its time de-
rivatives with respect to the starred system. Replacing dt*/dt, d 2 t*/dt in
Eqs. (7-5) and (7-6) by their expressions in terms of the starred deriva-
tives relative to the starred system as given by Eqs. (7-33) and (7-34),
we obtain for the position, velocity, and acceleration of a point with re-
spect to coordinate systems in relative translation and rotation :
r = r* + h, (7-38)
7-3 Laws of motion on the rotating earth. We write the equation of
motion, relative to a coordinate system fixed in space, for a particle of
mass ra subject to a gravitational force rag and any other nongravitational
forces F:
m^p=F + rag. (7-41)
Now if we refer the motion of the particle to a coordinate system at rest
relative to the earth, which rotates with constant angular velocity u, and
if we measure the position vector r from the center of the earth, we have,
by Eq. (7-34) :
d* 2 r d*i
= ra -jTg- + raw X (« X r) + 2mu X -3- » (7-42)
which can be rearranged in the form
d* 2 r d*r
m — ■ = F + ra[g - » X (w X r)] - 2ra« X ~- • (7-43)
7-3]
LAWS OF MOTION ON THE ROTATING EARTH
279
This equation has the same form as Newton's equation of motion. We
have combined the gravitational and centrifugal force terms because both
are proportional to the mass of the particle and both depend only on the
position of the particle; in their mechanical effects these two forces are
indistinguishable. We may define the effective gravitational acceleration
g e at any point on the earth's surface by:
g.(r) = g(r) - co X (co X r).
(7-44)
The gravitational force which we measure experimentally on a body of
mass m at restf on the earth's surface is mg e . Since — w X (co X r) points
radially outward from the earth's axis, g e at every point north of the
equator will point slightly to the south of the earth's center, as can be
seen from Fig. 7-5. A body released near the earth's surface will begin
to fall in the direction of g e , the direction determined by a plumb line is
that of g e , and a liquid will come to equilibrium with its surface perpen-
dicular to g e - This is why the earth has settled into equilibrium in the
form of an oblate ellipsoid, flattened at the poles. The degree of flattening
is just such as to make the earth's surface at every point perpendicular to
g e (ignoring local irregularities).
Equation (7-43) can now be written
d* 2 r d*T
(7-45)
The velocity and acceleration which appear in this equation are unaffected
if we relocate our origin of coordinates at any convenient point at the sur-
face of the earth; hence this equation applies to the motion of a particle of
mass m at the surface of the earth relative to a local coordinate system at
rest on the earth's surface. The only unfamiliar term is the coriolis force
« x (w x r)
Fig. 7-5. Effective acceleration of gravity on the rotating earth.
t A body in motion is subject also to the coriolis force.
280 MOVING COORDINATE SYSTEMS [CHAP. 7
which acts on a moving particle. The reader can convince himself by a
few calculations that this force is comparatively small at ordinary veloci-
ties d*i/dt. It will be instructive to try working out the direction of the
coriolis force for various directions of motion at various places on the
earth's surface. The coriolis force is of major importance in the motion
of large air masses, and is responsible for the fact that in the northern
hemisphere tornados and cyclones circle in the direction south to east to
north to west. In the northern hemisphere, the coriolis force acts to de-
flect a moving object toward the right. As the winds blow toward a low
pressure area, they are deflected to the right, so that they circle the low
pressure area in a counterclockwise direction. An air mass circling in this
way will have a low pressure on its left, and a higher pressure on its right.
This is just what is needed to balance the coriolis force urging it to the
right. An air mass can move steadily in one direction only if there is a
high pressure to the right of it to balance the coriolis force. Conversely,
a pressure gradient over the surface of the earth tends to develop winds
moving at right angles to it. The prevailing westerly winds in the northern
temperate zone indicate that the atmospheric pressure toward the equator
is greater than toward the poles, at least near the earth's surface. The
easterly trade winds in the equatorial zone are due to the fact that any air
mass moving toward the equator will acquire a velocity toward the west
due to the coriolis force acting on it. The trade winds are maintained by
high pressure areas on either side of the equatorial zone.
7-4 The Foucault pendulum. An interesting application of the theory
of rotating coordinate systems is the problem of the Foucault pendulum.
The Foucault pendulum has a bob hanging from a string arranged to swing
freely in any vertical plane. The pendulum is started swinging in a defi-
nite vertical plane and it is observed that the plane of swinging gradually
precesses about the vertical axis during a period of several hours. The
bob must be made heavy, the string very long, and the support nearly
frictionless, in order that the pendulum can continue to swing freely for
long periods of time. If we choose the origin of coordinates directly below
the point of support, at the point of equilibrium of the pendulum bob of
mass m, then the vector r will be nearly horizontal, for small amplitudes of
oscillation of the pendulum. In the northern hemisphere, a points in the
general direction indicated in Fig. 7-6, relative to the vertical. Writing t
for the tension in the string, we have as the equation of motion of the bob,
according to Eq. (7—15) :
d* 2 i d*i
m -^p- = r + mg e — 2mu X -^- • (7-46)
If the coriolis force were not present, this would be the equation for a
7-4]
THE FOUCAULT PENDULUM
281
mge
simple pendulum on a nonrotating
earth. The coriolis force is very
small, less than 0.1% of the gravita-
tional force if the velocity is 5 mi/hr
or less, and its vertical component is
therefore negligible in comparison
with the gravitational force. (It is
the vertical force which determines
the magnitude of the tension in the
string.) However, the horizontal
component of the coriolis force is
perpendicular to the velocity d*r/dt,
and as there are no other forces in
this direction when the pendulum
swings to and fro, it can change the
nature of the motion. Any force with a horizontal component perpendic-
ular to d*r/dt will make it impossible for the pendulum to continue to
swing in a fixed vertical plane. In order to solve the problem including
the coriolis term, we use the experimental result as a clue, and try to find
a new coordinate system rotating about the vertical axis through the
point of support at such an angular velocity that in this system the
coriolis terms, or at least their horizontal components, are missing. Let
us introduce a new coordinate system rotating about the vertical axis
with constant angular velocity kQ, where k is a vertical unit vector. We
shall call this precessing coordinate system the primed coordinate system,
and denote the time derivative with respect to this system by d'/dt.
Then we shall have, by Eqs. (7-33) and (7-34) :
Fig. 7-6. The Foucault pendulum.
d< 2 dl 2
Equation (7-46) becomes
d' 2 T
m ~M2 = T + m &°
d*I d'T , n . ^
J*2_ j/2_
= ^ + a 2 kx(kxr)+2!lkx
dt '
(7-47)
(7-48)
2m« X t ^ + Qk X i)
m£2 2 k X (k X r) — 2mGk X ^f
d't
dt
r + mg e — 2mO&> x (k x r) — mfi 2 k X (k X r)
d't
2m(w + kfi) X
dt
(7-49)
282 MOVING COORDINATE SYSTEMS [CHAP. 7
We expand the triple products by means of Eq. (3-35) :
d' 2 r
2i,
m-375- = t + mg e — m(2fiwr + fi k-r)k
<8 a
+ m(2fik-w + fi 2 )r - 2m(« + kO) X ^ • (7-50)
Every vector on the right side of Eq. (7-50) lies in the vertical plane con-
taining the pendulum, except the last term. Since, for small oscillations,
d'r/dt is practically horizontal, we can make the last term lie in this vertical
plane also by making (« + kfi) horizontal. We therefore require that
k-( w + kSi) = 0. (7-51)
This determines :
U = — w cos 0, (7-52)
where w is the angular velocity of the rotating earth, fi is the angular
velocity of the precessing coordinate system relative to the earth, and 6 is
the angle between the vertical and the earth's axis, as indicated in Fig. 7-6.
The vertical is along the direction of — g e , and since this is very nearly the
same as the direction of — g (see Fig. 7-5), 6 will be practically equal to the
colatitude, that is, the angle between r and w in Fig. 7-5. For small oscilla-
tions, if fi is determined by Eq. (7-52), the cross product in the last term
of Eq. (7-50) is vertical. Since all terms on the right of Eq. (7-50) now
lie in a vertical plane containing the pendulum, the acceleration d' 2 r/dt 2
of the bob in the precessing system is always toward the vertical axis, and
if the pendulum is initially swinging to and fro, it will continue to swing
to and fro in the same vertical plane in the precessing coordinate system.
Relative to the earth, the plane of the motion precesses with angular
velocity Q of magnitude and sense given by Eq. (7-52). In the northern
hemisphere, the precession is clockwise looking down.
Since the last three terms on the right in Eq. (7-50) are much smaller
than the first two, the actual motion in the precessing coordinate system
is practically the same as for a pendulum on a nonrotating earth. Even at
large amplitudes, where the velocity d'r/dt has a vertical component, care-
ful study will show that the last term in Eq. (7-50), when Q. is chosen
according to Eq. (7-52), does not cause any additional precession relative
to the precessing coordinate system, but merely causes the bob to swing in
an arc which passes slightly east of the vertical through the point of sup-
port. At the equator, Q. is zero, and the Foucault pendulum does not pre-
cess; by thinking about it a moment, perhaps you can see physically why
this is so. At the north or south pole, £2 = ±w, and the pendulum merely
swings in a fixed vertical plane in space while the earth turns beneath it.
7-5] lamor's theorem 283
Note that we have been able to give a fairly complete discussion of the
Foucault pendulum, by using Coriolis' theorem twice, without actually
solving the equations of motion at all.
7-5 Laimor's theorem. The coriolis force in Eq. (7-37) is of the same
form as the magnetic force acting on a charged particle (Eq. 3-281), in
that both are given by the cross product of the velocity of the particle
with a vector representing a force field. Indeed, in the general theory of
relativity, the coriolis forces on a particle in a rotating system can be re-
garded as due to the relative motion of other masses in the universe in a
way somewhat analogous to the magnetic force acting on a charged par-
ticle which is due to the relative motion of other charges. The similarity
in form of the two forces suggests that the effect of a magnetic field on a
system of charged particles may be canceled by introducing a suitable
rotating coordinate system. This idea leads to Larmor's theorem, which
we state first, and then prove:
Larmor's theorem. If a system of charged particles, all having the same
ratio q/m of charge to mass, acted on by their mutual (central) forces, and
by a central force toward a common center, is subject in addition to a weak
uniform magnetic field B, its possible motions will be the same as the motions
it could perform without the magnetic field, superposed upon a slow pre-
cession of the entire system about the center of force with angular velocity
- = ~ db B (7 " 53)
The definition of a weak magnetic field will appear as the proof is de-
veloped. We shall assume that all the particles have the same charge q
and the same mass m, although it will be apparent that the only thing that
needs to be assumed is that the ratio q/m is constant. Practically the
only important applications of Larmor's theorem are to the behavior of
an atom in a magnetic field. The particles here are electrons of mass m,
charge q = — e, acted upon by their mutual electrostatic repulsions and
by the electrostatic attraction of the nucleus.
Let the central force acting on the fcth particle be ¥%, and let the sum of
the forces due to the other particles be F\. Then the equations of motion
of the system of particles, in the absence of a magnetic field, are
m^i = F c k + Ft, k = 1, . . . , N, (7-54)
where N is the total number of particles. The force F% depends only on
the distance of particle fc from the center of force, which we shall take as
origin, and the forces F^ depend only on the distances of the particles from
284 MOVING COORDINATE SYSTEMS [CHAP. 7
one another. When the magnetic field is applied, the equations of motion
become, by Eq. (3-281) :
TO
% = n + n + l^xB, k=l,...,N. (7-55)
In order to eliminate the last term, we introduce a starred coordinate sys-
tem with the same origin, rotating about this origin with angular velocity
«. Making use of Eqs. (7-33) and (7-34), we can write the equations of
motion in the starred coordinate system:
m -^ = F| + Pi - ma X (a X i k ) + 2 (« x t t ) X B
We can make the last term vanish by setting
Equation (7-56) then becomes
m IP* = F * + F * + li B x < B x r *)> k=l,...,N. (7-58)
The forces F£ and F^ depend only on the distances of the particles from the
origin and on their distances from one another, and these distances will be
the same in the starred and unstarred coordinate systems. Therefore, if
we neglect the last term, Eqs. (7-58) have exactly the same form in
terms of starred coordinates as Eqs. (7-54) have in unstarred coordinates.
Consequently, their solutions will then be the same, and the motions of
the system expressed in starred coordinates will be the same as the motions
of the system expressed in unstarred coordinates in the absence of a mag-
netic field. This is Larmor's theorem.
The condition that the magnetic field be weak means that the last term
in Eq. (7-58) must be negligible in comparison with the first two terms.
Notice that the term we are neglecting is proportional to B 2 , whereas the
term in Eq. (7-55) which we have eliminated is proportional to B. Hence,
for sufficiently weak fields, the former may be negligible even though the
latter is not. The last term in Eq. (7-58) may be written in the form
2
-^ B x (B X r k ) = mu X (« X r»). (7-59)
Another way of formulating the condition for a weak magnetic field is to
7-6] THE RESTRICTED THREE-BODY PROBLEM 285
say that the Larmor frequency o>, given by Eq. (7-57), must be small com-
pared with the frequencies of the motion in the absence of a magnetic
field.
The reader who has understood clearly the above derivation should be
able to answer the following two questions. The cyclotron frequency,
given by Eq. (3-299), for the motion of a charged particle in a magnetic
field is twice the Larmor frequency, given by Eq. (7-57). Why does not
Larmor's theorem apply to the charged particles in a cyclotron? Equa-
tion (7-58) can be derived without any assumption as to the origin of co-
ordinates in the starred system. Why is it necessary that the axis of rota-
tion of the starred coordinate system pass through the center of force of
the system of particles?
7-6 The restricted three-body problem. We pointed out in Section 4-9
that the three-body problem, in which three masses move under their
mutual gravitational forces, cannot be solved in any general way. In this
section we will consider a simplified problem, the restricted problem of
three bodies, which retains many features of the more general problem,
among them the fact that there is no general method of solving it. In the
restricted problem, we are given two bodies of masses M\ and M 2 that
revolve in circles under their mutual gravitational attraction and around
their common center of mass. The third body of very small mass m moves
in the gravitational field of Mi and M 2 . We are to assume that m is so
small that the resulting disturbance of the motions of M i and M 2 can be
neglected. We will further simplify the problem by assuming that m
remains in the plane in which Mi and M 2 revolve. The problem thus
reduces to a one-body problem in which we must find the motion of m in
the given (moving) gravitational field of the other two. An obvious ex-
ample would be a rocket moving in the gravitational fields of the earth and
the moon, which revolve very nearly in circles about their common center
of mass.
If M\ and M 2 are separated by a distance a, then according to the
results of Section 4-7, their angular velocity is determined by equating the
gravitational force to mass times acceleration in the reduced problem, in
which Mi is at rest and M 2 has mass n as given by Eq. (4-98) :
so that
2 MiM 2 G (n Rm
u2= (Mi + M 2 )G
a 6
The center of mass divides the distance a into segments that are propor-
tional to the masses.
286 MOVING COORDINATE SYSTEMS [CHAP. 7
We now introduce a coordinate system rotating with angular velocity
w about the center of mass of M x and M 2 . In this system, M t and M 2 are
at rest, and we will take them to be on the ar-axis at the points
M 2 M\
Xl = M 1 + M 2 a > X2== ~M 1+ M 2 a - ff" 62 )
The angular velocity to is taken to be along the 2-axis. Then m moves in
the xy-pl&ne, and its equation of motion is
J*2 ,*
m -^ = Fx + F 2 - mu X (» X r) - 2ww X ~ > (7-63)
where F x and F2 are the gravitational attraction of Mi and M 2 on m.
Written in terms of components, the two equations become
M,G{x - Xl ) M 2 G(x - x 2 ) (Mj + M 2 )Gx .
[(x-a;i)a+y2]8/a [(3 _ X2 )2 + ^3/2 f a a T ^^>
.. = _ M x Gy M 2 Gy (M 1 + M 2 )Gy _ .
[(X - Xtf + 2/2J3/2 [( x _ X2 )2 +2/ 2]3/2^ a 3 ^"^
(7-64)
Note that the mass m cancels in these equations.
Since the coriolis force is perpendicular to the velocity, it does no
'work' in this moving coordinate system. Moreover, the centrifugal force
has zero curl and can be derived from the 'potential energy'
V c = -imcoV + y 2 ). (7-65)
Therefore the total 'energy' in the moving coordinate system is a constant
of the motion :
'E' = \m(x 2 + y 2 ) + 'V, (7-66)
where
, v , __ _ mM-iG mM 2 G
[{X — Zi)2 + 2/2]l/2 [( x — X2 )2 _|_ ^1/2
m(M 1 + M 2 )G(x 2 + y 2 )
2a,3
(7-67)
The energy equation (7-66) enables us to make certain statements about
the kinds of orbits that may be possible. In order to simplify the algebra,
let us set
£ = x/a, v = y/a, (7-68)
^ ~ MT+m' h= ~ Mi+M 2 = *i - !■ C 7 " 69 )
7-6] THE RESTRICTED THREE-BODY PROBLEM
Then Eq. (7-67) can be written as
m{M l + M 2 )G I £ 2
287
'V
Utt
fl )2 + ,2]l/2
Si
1
[(* - £a) 2 + >7 2 ] 1/2 2 ( * +
» 2 )}
(7-70)
In order to see the nature of this function, let us first look for its singular
points, where d'V'/d£ and d'V'/dy both vanish:
&(« - €i)
[(« - Si) 2 + » 2 ] 3/2
€2*1
+
Si CI - £2)
[(«
£ 2 )2 + ,2]3/2
Siq
K* - Si) 2 + v 2 ) 3 ' 2 ' [(S - S2) 2 + v 2 ] 312
+
— t\
s = o,
0.
(7-71)
A point Or, j/) for which these equations are satisfied is an equilibrium
point for the mass m (in the rotating coordinate system), since Eqs. (7-64)
are evidently satisfied if m is at rest at this point. We first consider points
on the rj = axis. The second equation is then satisfied, and the first
becomes
|S~S
Si) ■ Si(S - S2)
I q "T" 1 > > I a
S = 0.
(7-72)
In Fig. 7-7, we plot the function 'V, as given by Eq. (7-70), along the
ij = axis. The roots of Eq. (7-72) are the maxima of ' V'(f, 0) in Fig. 7-7,
where it can be seen that there are three such roots. Let us call them
£a, Sb> £c as in the figure. Each is the root of a quintic equation which
may be derived from Eq. (7-72). It is not difficult to show that
d 2 'V'/dt dr,= 0, d 2 'V'/dt 2 < 0, and d 2 'V'/dr, 2 > at these points A, B,
Fig. 7-7. A plot of 'F'(S, 0).
288
MOVING COORDINATE SYSTEMS
y
[chap. 7
Fig. 7-8. Equipotential contours for 'V'(x, y).
and C. If we expand 'V in a Taylor series about any one of these points,
and consider only the quadratic terms, we see that the curves of constant
'V are hyperbolas in the £ij-plane in the neighborhood of points A, B, C,
as shown in Fig. 7-8, where we plot the contours of constant 'V. These
points are saddlepoints of 'V; that is, 'V has a local maximum along the
{-axis and a minimum along a line perpendicular to the {-axis at each of
these points A, B, C. If i\ ?* 0, it can be factored from the second of
Eqs. (7-71). We then multiply the second of Eqs. (7-71) by (£ — |j)
and subtract from the first of these equations. After some manipulation
and using Eq. (7-69), we obtain
and, similarly,
(I - £ 2 ) 2 + v 2 = 1,
U - li) 2 + v 2 = l.
(7-73)
(7-74)
These equations show that there are two singular points D, E, off the
ij = axis, which lie at unit distance from (f i, 0) and (£ 2 , 0) which are
7-6] THE RESTRICTED THREE-BODY PROBLEM 289
themselves separated by a unit distance. By expanding 'V in a Taylor
series about point D or E, we can show that curves of constant 'V are
ellipses in the neighborhood of D or E, and that 'V has a maximum at D
and E. Knowing the behavior near the singular points, we can easily
sketch the general appearance of the contours of constant 'V, as shown
in Fig. 7-8. The curves are numbered in order of increasing 'V.
If this were a fixed coordinate system, we could immediately conclude
that equilibrium points A, B, C, D, E are all unstable, since the force
—V'V is directed away from each equilibrium point when m is at some
nearby points. However, this argument does not hold here because it
neglects the coriolis force in Eqs. (7-64). If we expand the right members
of Eqs. (7-64) in powers of the displacements (say x — xo,y — yo)
from one of the equilibrium points (say D), and retain only linear terms,
we may determine approximately the motion near the equilibrium point.
If this is done near point D (or E), for example, we find that in linear
approximation, the motion near D is stable if one of the masses M i or
M 2 contains more than about 96% of the total mass (Mi + M 2 ). (See
Problem 16.) For motions very near to point D, we may expect the linear
approximation to yield a solution which is valid for very long times.
Whether those motions which are stable in linear approximation are truly
stable, in the sense that they remain near point D for all time, is one of
the unsolved problems of classical mechanics. This matter is discussed
further at the end of Section 12-6.*
It is not difficult to show that, even in linear approximation, the equi-
librium points A, B,C are unstable. If the motion in linear approximation
is unstable, then the exact solution is certainly unstable. That is, regard-
less of how close m is initially to the equilibrium point (but not at it), it
will not, in general, remain as close but will move exponentially away, at
least at first. The neglected nonlinear terms may, of course, eventually
prevent the solution from going more than some finite distance from the
equilibrium point.
The only rigorous statements we can make about the motion of m, for
very long periods of time, are those which can be derived from the energy
equation (7-66). Given an initial position and velocity of m, we can
calculate 'E'. The orbit then must remain in the region where 'V < 'E'.
For example, motions which start near either mass Mi or M 2 , with
'E' < V 3 , must remain confined to a region near that mass. Motions
with 'E' > V5 may go to arbitrarily large distances; whether they actually
do, we cannot say from energy arguments. However, the studies which
* A more complete discussion of the problem of three bodies, on a more ad-
vanced level than the present text, will be found in Aurel Wintner, The Analytical
Foundations of Celestial Mechanics. Princeton: Princeton University Press, 1947.
290 MOVING COORDINATE SYSTEMS [CHAP. 7
have so far been made of the three-body problem make it very plausible,
though it has not been proved, that except for special cases (e.g., M 2 = 0)
or for special initial conditions, most orbits eventually wander throughout
the region that is 'energetically' allowed.
If we could find another constant of the motion, say F(x, y, x, y), we
could solve the problem by methods like those used in Chapter 3 for the
central force problem, where the angular momentum is also constant.
Unfortunately, no other such constant is known, and in view of the last
sentence of the preceding paragraph, it seems likely that none exists.
This problem has been studied very extensively.*
Faced with this situation, we may turn to the possibility of computing
particular orbits from given initial conditions. This can be done either
analytically, by approximation methods, or numerically, and in principle
can be done to any desired accuracy and for any desired finite period of
time.
In Chapter 12 we shall discuss a closely related special case of the three-
body problem.
Problems
1 . (a) Solve the problem of the freely falling body by introducing a translating
coordinate system with an acceleration g. Set up and solve the equations of
motion in this accelerated coordinate system and transform the result back to
a coordinate system fixed relative to the earth. (Neglect the earth's rotation.)
(b) In the same accelerated coordinate system, set up the equations of motion
for a falling body subject to an air resistance proportional to its velocity (rela-
tive to the fixed air).
2. A mass m is fastened by a spring (spring constant k) to a point of support
which moves back and forth along the z-axis in simple harmonic motion at
frequency oj, amplitude a. Assuming the mass moves only along the a>axis, set
up and solve the equation of motion in a coordinate system whose origin is at
the point of support.
3. Generalize Eq. (5-5) to the case when the origin of the coordinate system
is moving, by adding fictitious torques due to the fictitious force on each particle.
Express the fictitious torques in terms of the total mass M, the coordinate R*
of the center of mass, and ah. Compare your result with Eq. (4-25).
4. Derive a formula for d 3 A./dt 3 in terms of starred derivatives relative to a
rotating coordinate system.
5. Westerly winds blow from west to east in the northern hemisphere with an
average speed v. If the density of the air is p, what pressure gradient is required
to maintain a steady flow of air from west to east with this speed? Make reason-
able estimates of v and p, and estimate the pressure gradient in lb-in _2 -mile _1 .
: See A. Wintner, op. cit.
PROBLEMS 291
6. (a) It has been suggested that birds may determine their latitude by
sensing the coriolis force. Calculate the force a bird must exert in level flight
at 30 mi/hr against the sidewise component of coriolis force in order to fly in a
straight line. Express your result in g's, that is, as a ratio of coriolis force to
gravitational force, as a function of latitude and direction of flight.
(b) If the bird's flight path is slightly circular, a centrifugal force will be
present, which will add to the coriolis force and produce an error in estimated
latitude. At 45° N latitude, how much may the flight path bend, in degrees per
mile flown, if the latitude is to be determined within ±100 miles? (Assume the
sidewise force is measured as precisely as necessary!)
7. A body is dropped from rest at a height h above the surface of the earth,
(a) Calculate the coriolis force as a function of time, assuming it has a negligible
effect on the motion. Neglect air resistance, and assume h is small so that g e
can be taken as constant, (b) Calculate the net displacement of the point of
impact due to the calculated coriolis force.
*8. Find the answer to Problem 7(b) by solving for the motion in a non-
rotating coordinate system. What approximations are needed to arrive at the
same result?
9. A gyroscope consists of a wheel of radius r, all of whose mass is located on
the rim. The gyroscope is rotating with angular velocity 6 about its axis, which
is fixed relative to the earth's surface. We choose a coordinate system at rest
relative to the earth whose z-axis coincides with the gyroscope axis and whose
origin lies at the center of the wheel. The angular velocity u of the earth lies
in the zz-plane, making an angle a with the gyroscope axis.
Find the x-, y-, and z-components of the torque N about the origin, due to the
coriolis force in the xyz-coordinate system, acting on a mass m on the rim of the
gyroscope wheel whose polar coordinates in the zt/-plane are r, 6. Use this
result to show that the total coriolis torque on the gyroscope, if the wheel has
a mass M, is
N = jMr 2 u8 sin a.
This equation is the basis for the operation of the gyrocompass.
*10. A mass m of a perfect gas of molecular weight M, at temperature T, is
placed in a cylinder of radius a, height h, and whirled rapidly with an angular
velocity a about the axis of the cylinder. By introducing a coordinate system
rotating with the gas, and applying the laws of static equilibrium, assuming
that all other body forces are negligible compared with the centrifugal force,
show that
RT
V = -jf Po exp
/ Mo?r 2 \
\2RT )'
where p is the pressure, r is the distance from the axis, and
mMia
P0 = 2-rhRT[exp (Ma 2 a 2 /2RT) — 1] '
292 MOVING COORDINATE SYSTEMS [CHAP. 7
*11. A particle moves in the xy-pl&ne under the action of a force
F = — At,
directed toward the origin. Find its possible motions by introducing a coordinate
system rotating about the z-axis with angular velocity w chosen so that the
centrifugal force just cancels the force F, and solving the equations of motion
in this coordinate system. Describe the resulting motions, and show that your
result agrees with that of Problem 31, Chapter 3.
12. A ball of mass m slides without friction on a horizontal plane at the sur-
face of the earth. Show that it moves like the bob of a Foucault pendulum
of length equal to the earth's radius, provided it remains near the point of
tangency.
13. The bob of a pendulum is started so as to swing in a circle. By substitut-
ing in Eq. (7-46), find the angular velocity and show that the contribution due
to the coriolis force is given very nearly by Eq. (7-52). Neglect the vertical
component of the coriolis force, after showing that it is zero on the average for
the assumed motion.
14. An electron revolves about a fixed proton in an ellipse of semimajor axis
10 ~ 8 cm. If the corresponding motion occurs in a magnetic field of 10,000
gausses, show that Larmor's theorem is applicable, and calculate the angular
velocity of precession of the ellipse.
15. Write down a potential energy for the last term in Eq. (7-58). If the
plane of the orbit in Problem 14 is perpendicular to B, and if the orbit is very
nearly circular, calculate (by the methods of Chapter 3) the rate of precession
of the ellipse due to the last term in Eq. (7-58) in the rotating coordinate system.
Is this precession to be added to or subtracted from that calculated in Problem 14?
16. Find the three second derivatives of 'V with respect to £, ij for the point
D in Fig. 7-7. Expand the equations of motion (7-64), keeping terms linear in
£' = £ — £ and rj = i\ — ?;o. Using the method of Section 4-10, find the
condition on Mi, M% in order that the normal modes of oscillation be stable.
If M\ > M.2, what is the minimum value of Mi/ (Mi + M2)?
17. Prove the statements made in Section 7-6 regarding the second deriva-
tives of 'V at points A, B, and C in Fig. 7-7. Expand the equations of motion
about points A and B, keeping terms linear in jj and £' = £ — £a,b- Show by
the method of Section 4-10 that some of the solutions are unstable for any
values of the masses. (You cannot find the second derivatives explicitly, but the
proof depends only on their signs.)
*18. (a) Write out the quintic equation which must be solved for iU in Fig. 7-7.
Show that if M 2 = 0, the solution is % A = —1- (b) Solve numerically for £ A
to two decimal places for the earth-moon system, (c). Find the minimum
launching velocity from the surface of the earth for which it is 'energetically'
possible for a rocket to leave the earth-moon system. Compare with the escape
velocity from the earth.
19. Two planets, each of mass M and radius R, revolve in circles about each
other at a distance a apart. Find the minimum velocity with which a rocket
might leave one planet to arrive at the other. Show that the rocket must have a
PROBLEMS 293
larger velocity than would be calculated if the motion of the planets were neg-
lected.
20. (a) Locate all fixed points in the limiting case M 2 — » 0, and sketch Fig. 7-7
for this case. Show that the results in Section 7-6 applied to this case are con-
sistent with the complete solution given in Section 3-14.
(b) Show from this example for which the complete solution is known, that
the minimum 'energetically' possible launching velocity for escape calculated
as in Problem 18(c) is not necessarily the true minimum escape velocity.
CHAPTER 8
INTRODUCTION TO THE MECHANICS
OF CONTINUOUS MEDIA
In this chapter we begin the study of the mechanics of continuous
media, solids, fluids, strings, etc. In such problems, the number of parti-
cles is so large that it is not practical to study the motion of individual
particles, and we instead regard matter as continuously distributed in
space and characterized by its density. We are interested primarily in
gaining an understanding of the concepts and methods of treatment which
are useful, rather than in developing in detail methods of solving practical
problems. In the first four sections, we shall treat the vibrating string,
using concepts which are a direct generalization of particle mechanics. In
the remainder of the chapter, the mechanics of fluids will be developed in
a way less directly related to particle mechanics.
8-1 The equation of motion for the vibrating string. In this section we
shall study the motion of a string of length I, stretched horizontally and
fastened at each end, and set into vibration. In order to simplify the
problem, we assume the string vibrates only in a vertical plane, and that
the amplitude of vibration is small enough so that each point on the string
moves only vertically, and so that the tension in the string does not change
appreciably during the vibration.
We shall designate a point on the string by giving its horizontal distance
x from the left-hand end (Fig. 8-1). The distance the point x has moved
from the horizontal straight line representing the equilibrium position of
the string will be designated by u(x). Thus any position of the entire
string is to be specified by specifying the function u(x) for < x < I.
This is precisely analogous, in the case of a system of N particles, to speci-
fying the coordinates x t , ?/,-, z», for i = 1, . . . , N. In the case of the
string, x is not a coordinate, but plays the same role as the subscript i; it
designates a point on the string. Our idealized continuous string has
Fig. 8-1. The vibrating string.
294
8-1] THE EQUATION OF MOTION FOR THE VIBRATING STRING 295
infinitely many points, corresponding to the infinitely many values of x
between and I. For a given point x, it is u{x) that plays the role of a
coordinate locating that point, in analogy with the coordinates x iy yi, z t of
particle i. Just as a motion of the system of particles is to be described
by functions Xi(t), Vi(t), Zi(t), locating each particle at every instant of
time, so a motion of the string is to be described by a function u{x, t),
locating each point x on the string at every instant of time.
In order to obtain an equation of motion for the string, we consider a
segment of string of length dx between x and x + dx. If the density of the
string per unit length is <r, then the mass of this segment is <r dx. The
velocity of the string at any point is du/dt, and its slope is du/dx. The
vertical component of tension exerted from right to left across any point in
the string is
r„ = t sin 6, (8-1)
where 6 is the angle between the string and the horizontal (Fig. 8-1). We
are assuming that 6 is very small and, in this case,
t sin 6 = r tan = t — • (8-2)
The net upward force dF due to the tension, on the segment dx of string,
is the difference in the vertical component r„ between the two ends of the
segment :
dF = [r M ] x+ dx — [tu]x
(8-3)
dx \ dx/
If we do not limit ourselves to very small slopes du/dx, then a segment of
string may also have a net horizontal component of force due to tension,
and the segment will move horizontally as well as vertically, a possibility
we wish to exclude. If there is, in addition, a vertical force / per unit
length, acting along the string, the equation of motion of the segment dx
will be
d 2 u d
VlTx)
<rdx W = dx'\ T ^l dx+fdx - (8_4)
For a horizontal string acted on by no horizontal forces except at its ends,
and for small amplitudes of vibration, the tension is constant, and
Eq. (8-4) can be rewritten:
d 2 u d 2 u , . , a _.
296 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
The force / may be the gravitational force acting on the string, which is
usually negligible unless the tension is very small. The force / may also
represent an external force applied to the string to set it into vibration.
We shall consider only the case/ = 0, and we rewrite Eq. (8-5) in the form
d 2 u 1 d 2 u , .
ex*-T>W = > (8_6)
where
=(r-
(8-7)
The constant c has the dimensions of a velocity, and we shall see in Sec-
tion 8-3 that it is the velocity with which a wave travels along the string.
Equation (8-6) is a partial differential equation for the function u(x, t) ;
it is the mathematical expression of Newton's law of motion applied to
the vibrating string. We shall want to find solutions u(x, t) to Eq. (8-6),
for any given initial position u (x) of the string, and any given initial
velocity v {x) of each point along the string. If we take the initial instant
at t = 0, this means that we want a solution u(x, t) which satisfies the
initial conditions:
u(x, 0) = u (x),
(8-8)
f-1 =
v (x).
The solution must also satisfy the boundary conditions:
«(0, t) = u(l, t) = 0, (8-9)
which express the fact that the string is tied at its ends. From the nature
of the physical problem, we expect that there should be just one solution
u(x, t) of Eq. (8-6) which satisfies Eqs. (8-8) and (8-9), and this solution
will represent the motion of the string with the given initial conditions.
It is therefore reasonable to expect that the mathematical theory of partial
differential equations will lead to the same conclusion regarding the num-
ber of solutions of Eq. (8-6), and indeed it does.
8-2 Normal modes of vibration for the vibrating string. We shall first
try to find some solutions of Eq. (8-6) which satisfy the boundary condi-
tions (8-9), without regard to the initial conditions (8-8). This is analo-
gous to our treatment of the harmonic oscillator, in which we first looked
for solutions of a certain type and later adjusted these solutions to fit the
initial conditions of the problem. The method of finding solutions which
8-2] NORMAL MODES OF VIBRATION FOR THE VIBRATING STRING 297
we shall use is called the method of separation of variables. It is one of
the few general methods so far devised for solving partial differential equa-
tions, and many important equations can be solved by this method. Un-
fortunately, it does not always work. In principle, any partial differential
equation can be solved by numerical methods, but the labor involved in
doing so is often prohibitive, even for the modern large-scale automatic
computing machines.
The method of separation of variables consists in looking for solutions
of the form
u(x, t) = X(z)0(<), (8-10)
that is, u is to be a product of a function X of x and a function © of t. The
derivatives of u will then be
d 2 u „d 2 X d 2 u v d 2 @
w = @ !xT' W = X HP- < 8 - n)
If these expressions are substituted in Eq. (8-6), and if we divide through
by ©X, then Eq. (8-6) can be rewritten :
ld 2 X_l_d 2 ®. ~
X dx* _ dP (8_12)
The left member of this equation is a function only of x, and the right
member is a function only of t. If we hold t fixed and vary x, the right
member remains constant, and the left member must therefore be inde-
pendent of x. Similarly, the right member must actually be independent
of t. We may set both members equal to a constant. It is clear on physical
grounds that this constant must be negative, for the right member of
Eq. (8-12) is the acceleration of the string divided by the displacement, and
the acceleration must be opposite to the displacement or the string will not
return to its equilibrium position. We shall call the constant -co 2 :
1 d 2 ® 2 c 2 d 2 X a
*r = -" ' l¥=- w ' < 8 " 13 >
The first of these equations can be rewritten as
d 2 %
dP
+ o J e = 0, (8-14)
which we recognize as the equation for the harmonic oscillator, whose gen-
eral solution, in the form most suitable for our present purpose, is
© = A cos wt + B sin wt, (8-15)
298 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
where A and B are arbitrary constants. The second of Eqs. (8-13) has a
similar form :
d 2 X -- 2
dx 2
and has a similar solution :
+ ^- X = 0, (8-16)
X-Ccos— + Z>sin — ■ (8-17)
c c
The boundary condition (8-9) can hold for all times t only if X satisfies
the conditions
X(0) = C = 0,
X(l) = C cos - + D sin - = 0. (8-18)
c c
The first of these equations determines C, and the second then requires
that
sin - = 0. (8-19)
c
This will hold only if w has one of the values
Wn = 2p , n= 1, 2, 3, ... . (8-20)
Had we taken the separation constant in Eqs. (8-13) as positive, we would
have obtained exponential solutions in place of Eq. (8-17), and it would
have been impossible to satisfy the boundary conditions (8-18).
The frequencies v n = o) n /2ir given by Eq. (8-20) are called the normal
frequencies of vibration of the string. For a given n, we obtain a solution by
substituting Eqs. (8-15) and (8-17) in Eq. (8-10), and making use of Eqs.
(8-18), (8-20):
... . . nirx nirct , „ . nwx . rnrct , a 01 ,
u{x, t) = A sin — j- cos — j \- B sin — r— sin — j- > (8-21)
where we have set D = 1. This is called a normal mode of vibration of
the string, and is entirely analogous to the normal modes of vibration
which we found in Section 4-10 for coupled harmonic oscillators. Each
point on the string vibrates at the same frequency u n with an amplitude
which varies sinusoidally along the string. Instead of two coupled oscil-
lators, we have an infinite number of oscillating points, and instead of two
normal modes of vibration, we have an infinite number.
The initial position and velocity at t = of the nth normal mode of
vibration as given by Eq. (8-21) are
8-2] NORMAL MODES OF VIBRATION FOR THE VIBRATING STRING 299
Uo\%) = A SID. -y-
, s nircB . nirx
v (x) = — z — sin —j- ■
T'
(8-22)
Only for these very special types of initial conditions will the string vibrate
in one of its normal modes. However, we can build up more general solu-
tions by adding solutions; for the vibrating string, like the harmonic oscil-
lator, satisfies a principle of superposition. Let Ui(x, t) and u 2 (x, t) be any
two solutions of Eq. (8-6) which satisfy the boundary conditions (8-9).
Then the function
u(x, t) = «i (#, + U2(x, t)
also satisfies the equation of motion and the boundary conditions. This
is readily verified simply by substituting u{x, t) in Eqs. (8-6) and (8-9),
and making use of the fact that u x {x, t) and u 2 (x, t) satisfy these equations.
A more general solution of Eqs. (8-6) and (8-9) is therefore to be obtained
by adding solutions of the type (8-21), using different constants A and B
for each normal frequency :
u(x, t) = 2~i\ An sin ~T~ cos — z 1" B n sm ~T Sln ~T~ ) ' (8-23)
The initial position and velocity for this solution are
«oW = Zj An Sln ~T~ '
n=l *
,, , v-\ rnrcB n . nirx
Vo(x) = 2^ — i — Sln "J" -
(8-24)
n=l
Whether or not Eq. (8-23) gives a general solution to our problem depends
on whether, with suitable choices of the infinite set of constants A n , B n ,
we can make the functions u (x) and v (x) correspond to any possible
initial position and velocity for the string. Our intuition is not very clear
on this point, although it is clear that we now have a great variety of possi-
ble functions WoO^) and v (x). The answer is provided by the Fourier series
theorem, which states that any continuous function u (x) for (0 < x < I),
which satisfies the boundary conditions (8-9), can be represented by the
sum on the right in Eq. (8-24), if the constants A n are properly chosen.*
* R. V. Churchill, Fourier Series and Boundary Value Problems. New York:
McGraw-Hill, 1941. (Pages 57-70.) Even functions with a finite number of dis-
continuities can be represented by Fourier series, but this point is not of great
interest in the present application.
300 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
Similarly, with the proper choice of the constants B n , any continuous func-
tion v (x) for (0 < x < I) can be represented.* The expressions for A n
and B n are, in this case,
r i
2 -
IJo
r i
nirc Jo
A„ = j J u {x) sin —j- dx,
1 (8-25)
"» = ZZZ I "oW Sln —y~ ® x -
The most general motion of the vibrating string is therefore a superposi-
tion of normal modes of vibration at the fundamental frequency V\ = c/2l
and its harmonics v n = nc/2l.
8-3 Wave propagation along a string. Equations (8-14) and (8-16)
have also the complex solutions
= Ae ±iat , (8-26)
x = gi'O/o)*. (8-27)
Hence Eq. (8-6) has complex solutions of the form
u(x, t) = Ae ±iialcnx±ct \ (8-28)
By taking the real part, or by adding complex conjugates and dividing
by 2, we obtain the real solutions
u(x, t) = A cos - (x - ct), (8-29)
c
u(x, t) = A cos - (x + ct). (8-30)
By taking imaginary parts, or by subtracting complex conjugates and
dividing by 2i, we could obtain similar solutions with cosines replaced by
sines. These solutions do not satisfy the boundary conditions (8-9), but
they are of considerable interest in that they represent waves traveling
down the string, as we now show.
A fixed point x on the string will oscillate harmonically in time, accord-
ing to the solution (8-29) or (8-30), with amplitude A and angular fre-
quency a. At any given instant t, the string will be in the form of a sinus-
* The Fourier series theorem was quoted in Section 2-11 in a slightly different
form. The connection between Eqs. (8-24) and (2-205) is to be made by replacing
t by x and T by 21 in Eq. (2-205). Both sine and cosine terms are then needed to
represent an arbitrary function uo(x) in the interval (0 < x < 21), but only sine
terms are needed if we want to represent uo(x) only in the interval (0 < x < I).
[Cosine terms alone would also do for this interval, but sine terms are appropriate
if uo(x) vanishes at x = and x = I.]
8-3] WAVE PROPAGATION ALONG A STRING 301
oidal curve with amplitude A and wavelength X (distance between successive
maxima) :
X = ^- (8-31)
We now show that this pattern moves along the string with velocity c, to
the right in solution (8-29), and to the left in solution (8-30). Let
£ = x — ct, (8-32)
so that Eq. (8-29) becomes
u = A cos — » (8-33)
where { is called the phase of the wave represented by the function u. For
a fixed value of £ , u has a fixed value. Let us consider a short time interval
dt and find the increment dx required to maintain a constant value of |:
d£ = dx — c dt = 0. (8-34)
Now if dx and dt have the ratio given by Eq. (8-34),
dx
dt
= c, (8-35)
then the value of u at the point x + dx at time t + dt will be the same as
its value at the point x at time t. Consequently, the pattern moves along
the string with velocity c given by Eq. (8-7). The constant c is the phase
velocity of the wave. Similarly, the velocity dx/dt for solution (8-30) is — c.
It is often convenient to introduce the angular wave number k defined by
the equation
where k is taken as positive for a wave traveling to the right, and negative
for a wave traveling to the left. Then both solutions (8-29) and (8-30)
can be written in the symmetrical form
u = A cos (kx — «<)• (8-37)
The angular wave number k is measured in radians per centimeter, just as
the angular frequency « is measured in radians per second. The expres-
sion for u in Eq. (8-37) is the real part of the complex function
u = Ae i(kx - at \ (8-38)
This form is often used in the study of wave motion.
302 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
The possibility of superposing solutions of the form (8-29) and (8-30)
with various amplitudes and frequencies, together with the Fourier series
theorem, suggests a more general solution of the form
u(x, t) = f(x - ct) + g(x + ct), (8-39)
where /(£) and g(y) are arbitrary functions of the variables £ = x — ct,
and q = x + ct. Equation (8-39) represents a wave of arbitrary shape
traveling to the right with velocity c, and another traveling to the left.
We can readily verify that Eq. (8-39) gives a solution of Eq. (8-6) by
calculating the derivatives of u:
dx _ dt, dx dr, dx di dr, '
d 2 u __ d 2 f d 2 g
dx 2 ~~ d? + dr, 2 '
du = dfdl.dgih = _ d[ dg }
dt d£ dt dv dt d£ dri '
d 2 u _ ,dV 2 <Pg
dt 2 C dp dr, 2 '
When these expressions are substituted in Eq. (8-6), it is satisfied iden-
tically, no matter what the functions /(f) and g(rj) may be, provided, of
course, that they have second derivatives. Equation (8-39) is, in fact, the
most general solution of the equation (8-6) ; this follows from the theory
of partial differential equations, according to which the general solution
of a second-order partial differential equation contains two arbitrary func-
tions. We can prove this without resorting to the theory of partial differ-
ential equations by assuming the string to be of infinite length, so that
there are no boundary conditions to concern us, and by supposing that the
initial position and velocity of all points on the string are given by the
functions u (x), v {x). If the solution (8-39) is to meet these initial con-
ditions, we must have, at t — :
«(*, 0) = Six) + g(x) = u (x), (8-40)
At t — 0, £ = i\ = x, so that Eq. (8-41) can be rewritten :
£[-f(x)+g(.x)) = "-*&, (8-42)
8-3] WAVE PROPAGATION ALONG A STRING 303
which can be integrated to give
rx
-/(*) + g(x) = - / v (x) dx + C. (8-43)
C J
By adding and subtracting Eqs. (8-40) and (8-43), we obtain the functions
/andg:
f(x) = i\u (x) / v (x) dx — C),
9(x) = i( u (x) + - J v (x) dx + Cj-
(8-44)
The constant C can be omitted, since it will cancel out in u = f + g, and
we can replace x by £ and ij respectively in these equations :
/(*) = i(«ott) -\j o »o«)dt)
(8-45)
This gives a solution to Eq. (8-6) for any initial position and velocity of
the string.
Associated with a wave
u = f{x — ct), (8-46)
there is a flow of energy down the string, as we show by computing the
power delivered from left to right across any point x on the string. The
power P is the product of the upward velocity of the point x and the up-
ward force [Eq. (8-2)] exerted by the left half of the string on the right
half across the point x:
„ du du ,„ ,_.
P =- T 6x-di- <W
If u is given by Eq. (8-46), this is
P = cr(|)\ (8-48)
which is always positive, indicating that the power flow is always from
left to right for the wave (8-46). For a wave traveling to the left, P will
be negative, indicating a flow of power from right to left. For a sinusoidal
wave given by Eq. (8-37), the power is
P = IccotA 2 sin 2 (fee — at), (8-49)
or, averaged over a cycle,
(P)„ = ±kwrA 2 . (8-50)
304 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
We now consider a string tied at x = and extending to the left from
a; = to a; = — oo. The solution (8-39) must now satisfy the boundary
condition
«(0,0 = f(-ct) + y(ct) = 0, (8-51)
or
/(-*) = -9(S), (8-52)
for all values of £. The initial values uo(x) and v (x) will now be given
only for negative values of x, and Eqs. (8-45) will define /(£) and g(ri) only
for negative values of | and »j. The values of /(£) and g(rj) for positive
values of £ and jj can then be found from Eq. (8-52) :
/(I) = -*(-*), gin) = -/(-i). (8-53)
Let us consider a wave represented by f(x — ct) traveling toward the end
x = 0. A particular phase £o> for which the wave amplitude is/(£o)> will
at time t be at the point
x = So + ct . (8-54)
Let us suppose that £ and t are so chosen that x is negative. At a later
time <i, the phase £ will be at the point
xi = So + c<i = cc + c(«i — t ). (8-55)
At <i = t Q — (x /c), Xi = and the phase f reaches the end of the
string. At later times x x will be positive, and f{x x — ct{) will have no
physical meaning, since the string does not extend to positive values of x.
Now consider the phase i) » of the leftward traveling wave g{x + ct),
denned by
jjo = x + ct = — £ o- (8-56)
The amplitude of the leftward wave g(iio) f° r the phase ij is related to the
amplitude of the rightward wave/(£ ) for the corresponding phase £ by
Eq. (8-53):
fOlo) = -/(«o). (8-57)
At time ti, the phase ij will be at the point
X2 = Vo — c&i = —x — c(h — t ). (8-58)
If (<x — t ) > —x /c, x 2 is negative and 0(170) represents a wave of equal
and opposite amplitude to /(£o), traveling to the left. Thus the wave
fix — ct) is reflected out of phase at x = and becomes an equal and
opposite wave traveling to the left. (See Fig. 8-2.) The total distance
8-4] STRING AS LIMITING CASE OF SYSTEM OF PARTICLES 305
t = h
Fig. 8-2. A wave reflected at x = 0.
traveled by the wave during the time (ii — t ), from x = x to x =
and back to x = x 2 is, by Eq. (8-58),
— x — x 2 = c(«i — t ), (8-59)
as it should be.
The solution (8-39) can also be fitted to a string of finite length fastened
at x = and x = I. In this case, the initial position and velocity u (x)
and v (x) are given only for (0 < x < I). The functions /(£) and g(t)) are
then defined by Eq. (8-45) only for (0 < £ < I, < i? < I). If we define
/(|) and g(v) for negative values of £ and ij by Eq. (8-53), in terms of their
values for positive { and t\, then the boundary condition (8-51) will be satis-
fied at x = 0. By an argument similar to that which led to Eq. (8-53),
we can show ;hat the boundary condition (8-9) for x — I will be satisfied if,
for all value* of £ and t\,
/(£ + = -gQ-i),
(8-60)
9(V + I) = -fd -v).
By means of Eqs. (8-53) and (8-60), we can find/(£) and g{i\) for all values
of £ and v, once their values are given [by Eqs. (8-45)] for < £ < I,
< i\ < I. Thus we find a solution for the vibrating string of length I hi
terms of waves traveling in opposite directions and continuously being
reflected at x = and x — I. The solution is equivalent to the solution
given by Eqs. (8-23) and (8-25) in terms of standing sinusoidal waves.
8-4 The string as a limiting case of a system of particles. In the first
three sections of this chapter, we have considered an idealized string char-
acterized by a continuously distributed mass with density a and tension t.
An actual string is made up of particles (atoms and molecules) ; our treat-
ment of it as continuous is valid because of the enormously large number of
particles in the string. A treatment of an actual string which takes into
account the individual atoms would be hopelessly difficult, but we shall
consider in this section an idealized model of a string made up of a finite
number of particles, each of mass m. Figure 8-3 shows this idealized
string, in which an attractive force r acts between adjacent particles
306
THE MECHANICS OP CONTINUOUS MEDIA
[CHAP. 8
Fig. 8-3. A string made up of particles.
along the line joining them. The interparticle forces are such that in
equilibrium the string is horizontal, with the particles equally spaced a
distance h apart. The string is of length (N + l)h, with N + 2 particles,
the two end particles being fastened at the x-axis. The N particles which
are free to move are numbered 1,2, ... ,N, and the upward displacement
of particle j from the horizontal axis will be called Uj. It will be assumed
that the particles move only vertically and that only small vibrations are
considered, so that the slope of the string is always small. Then the
equations of motion of this system of particles are
m
d uj
dt 2
. u )+i
h
Uj Uj My — 1
j= 1,...,N, (8-61)
where the expression on the right represents the vertical components of the
forces r between particle j and the two adjacent particles, and we are sup-
posing that the forces r are equal between all pairs of particles. Now let
us assume that the number N of particles is very large, and that the dis-
placement of the string is such that at any time t, a smooth curve u(x, t)
can be drawn through the particles, so that
u(jh, t) = Uj{t).
(8-62)
We can then represent the system of particles approximately as a contin-
uous string of tension t, and of linear density
<T = -r
m
h '
The equations of motion (8-61) can be written in the form
d 2 Uj _ T 1 ( Uj + i — Uj _ Uj — Uj-j \
dt 2 ~ <r h\ h h ) '
(8-63)
(8-64)
Now if the particles are sufficiently close together, we shall have, approxi-
mately,
Uj+i — uj _^_ \du\
h Ldx] x =(j+l/2)h
Uj — Uj—i j_ dw
h L.dxXc=(j-ll2)h
(8-65)
8-4] STRING AS LIMITING CASE OF SYSTEM OF PARTICLES 307
and hence / \ r, 2 i
1 ( ttj+i — «; u i — u i-i ) ^. \ d u \ . C8-66')
h\ h h ) ldx 2 ] x=jh v ;
The function u(x, I) therefore, when h is very small, satisfies the equation
d u T a u /0 „-n
dt 2 a dx 2
which is the same as Eq. (8-6) for the continuous string.
The solutions of Eqs. (8-61) when JV is large will be expected to approxi-
mate the solutions of Eq. (8-6). If we were unable to solve Eq. (8-6)
otherwise, one method of solving it numerically would be to carry out the
above process in reverse, so as to reduce the partial differential equation
(8-67) to the set of ordinary differential equations (8-61), which could then
be solved by numerical methods. The solutions of Eqs. (8-61) are of
some interest in their own right. Let us rewrite these equations in the
form
m 1& + T U * ~ I {Uj+1 + Uj ~ l] = °' j = *' • • • ' N - (8_68)
These are the equations for a set of harmonic oscillators, each coupled to
the two adjacent oscillators. We are led, either by our method of treat-
ment of the coupled oscillator problem or by considering our results for the
continuous string, to try a solution of the form
Uj = aje ±iat . (8-69)
If we substitute this trial solution in Eqs. (8-68), the factor e ±iat cancels
out, and we get a set of algebraic equations:
(2t ,\ t t
-j — mw J a, — ^ a i+ i — ^ a y _i = 0, j = 1, . . . , N. (8-70)
This is a set of linear difference equations which could be solved for o J+ i
in terms of a, and o,_i. Since a = 0, if a± is given we can find the values
of the remaining constants a,- by successive applications of these equations.
A neater method of solution is to notice the analogy between the linear
difference equations (8-70) and the linear differential equation (8-16),
and to try the solution
aj = Ae ipj , j=l,...,N. (8-71)
When this is substituted in Eqs. (8-70), we get, after canceling the factor
Ae ip >:
(?1 - mJ) - \ ie iv + e~ ip ) = 0, (8-72)
308 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
OT J, 2
cos p = 1 — • (8-73)
If a is less than
\mh/
(8-74)
there will be real solutions for p. Let a solution be given by
p = kh, < kh < ir. (8-75)
Then another solution is
V = —kh. (8-76)
All other solutions for p differ from these by multiples of 27r, and in view
of the form of Eq. (8-71), they lead to the same values of ay, so we can re-
strict our attention to values of p given by Eqs. (8-75) and (8-76).
If we substitute Eq. (8-71) in Eq. (8-69), making use of Eq. (8-75), we
have a solution of Eqs. (8-68) in the form
Uj = Ae± iikik -"\ (8-77)
Since the horizontal distance of particle j from the left end of the string is
Xj = jh, (8-78)
we see that the solution (8-77) corresponds to our previous solution (8-38)
for the continuous string, and represents traveling sinusoidal waves. By
combining the two complex conjugate solutions (8-77) and using Eq.
(8-78), we obtain the real solution
My = A cos (kx, — (tit), (8-79)
which corresponds to Eq. (8-37). We thus have sinusoidal waves which
may travel in either direction with the velocity [Eq. (8-36)]
u ha ._ „„.
where p is given by Eq. (8-73). If u <3C u c [Eq. (8-74)], then p will be
nearly zero, and we can expand cos p in Eq. (8-73) in a power series:
t p 2 . mhu> 2
-(?)'"•
|p|=col — ) , (8-81)
8-4] STRING AS LIMITING CASE OF SYSTEM OF PARTICLES 309
and
*(£)"■•
(8-82)
which agrees with Eq. (8-7) for the continuous string, in view of Eq.
(8-63). However, for larger values of u, the velocity c is smaller than for
the continuous string, and approaches
c^jfcf (8-83)
7T IT \m/
as w — * w c . (w c = oo for the continuous string for which mh = ah 2 = 0.)
Since the phase velocity given by Eq. (8-80) depends upon the frequency,
we cannot superpose sinusoidal solutions to obtain a general solution of the
form (8-39). If a wave of other than sinusoidal shape travels along the
string, the sinuosidal components into which it may be resolved travel with
different velocities, and consequently the shape of the wave changes as it
moves along. This phenomenon is called dispersion.
When (>) > w e , Eq. (8-73) has only complex solutions for p, of the form
p = ir ± i7. (8-84)
These lead to solutions Uj of the form
Uj = (- lYAe ±yj cos ut. (8-85)
There is then no wave propagation, but only an exponential decline in
amplitude of oscillation to the right or to the left from any point which
may be set in oscillation. The minimum wavelength [Eq. (8-36)] which
is allowed by Eq. (8-75) is
X c = f£ = 2h. (8-86)
It is evident that a wave of shorter wavelength than this would have no
meaning, since there would not be enough particles in a distance less than
X c to define the wavelength. The wavelength X c corresponds to the fre-
quency w c , for which
Uj = Ae iJT e ±ia ° l = (-lj^e**"'. (8-87)
Adjacent particles simply oscillate out of phase with amplitude A.
We can build up solutions which satisfy the boundary conditions
« = u N+ i = (8-88)
by adding and subtracting solutions of the form (8-77). We can, by suit-
ably combining solutions of the form (8-77), obtain the solutions
310 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
Uj = A sin pj cos ut + B sin pj sin cat + C cos pj cos cot
+ D cos pj sin wf . (8-89)
In order to satisfy the conditions (8-88), we must set
C = D = 0,
TITT
P = j\m ' n=l,2,...,N, (8-90)
where the limitation n < N arises from the limitation on p in Eq. (8-75).
The normal frequencies of vibration are now given by Eq. (8-73) :
Cn = [^V~ C0S WTv\ '
n = 1, 2, . . . , N. (8-91)
If n <<C N, we can expand the cosine in a power series, to obtain
Wn ~ lmh(N -
h{N + 1) 2 J
which agrees with Eq. (8-20) for the continuous string.
A physical model which approximates fairly closely the string of particles
treated in this section can be constructed by hanging weights m at intervals
h along a stretched string. The mass m of each weight must be large in
comparison with that of a length h of the string.
8-5 General remarks on the propagation of waves. If we designate
by F the upward component of force due to tension, exerted from left to
right across any point in a stretched string, and by v the upward velocity
of any point on the string, then, by Eq. (8-2), we have
By Eq. (8-4), if there is no other force on the string, we have
Tt = -*te' (8 ' 95)
and by differentiating Eq. (8-93) with respect to t, assuming t is constant
8-5] GENERAL REMARKS ON THE PROPAGATION OF WAVES 311
in time, we obtain „
-^-r-. (8-96)
Equations (8-95) and (8-96) are easily understood physically. The ac-
celeration of the string will be proportional to the difference in the upward
force F at the ends of a small segment of string. Likewise, since F is pro-
portional to the slope, the time rate of change of F will be proportional to
the difference in upward velocities of the ends of a small segment of the
string. The power delivered from left to right across any point in the
string is
P = Fv. (8-97)
Equations (8-95) and (8-96) are typical of many types of small ampli-
tude wave propagation which occur in physics. There are two quantities,
in this case F and v, such that the time rate of change of either is propor-
tional to the space derivative of the other. For large amplitudes, the equa-
tions for wave propagation may become nonlinear, and new effects like the
development of shock fronts may occur which are not described by the
equations we have studied here. When there is dispersion, linear terms in
v and F, or terms involving higher derivatives than the first, may appear.
Whenever equations of the form (8-95) and (8-96) hold, a wave equation
of the form (8-6) can be derived for either of the two quantities. For ex-
ample, if we differentiate Eq. (8-95) with respect to t, and Eq. (8-96)
with respect to x, assuming a, t to be constant, we obtain
or
dx 2 c 2 dt 2
d 2 F
dxdt
d 2 v
T dx 2
d 2 v
,2
d V
1 d 2 v
where
<r-
0, (8-98)
(8-99)
Similarly, we can show that
Usually one of these two quantities can be chosen so as to be analogous to
a force (F), and the other to the corresponding velocity (v), and then the
power transmitted will be given by an equation like Eq. (8-97). Likewise,
all other quantities associated with the wave motion satisfy a wave equa-
tion, as, for example, u, which satisfies Eq. (8-6).
312 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
As a further example, the equations for a plane sound wave traveling in
the z-direction, which will be derived in Section 8-10, can be written in
the form
dv _ I dp' dp' _ dv
di--p~-dx-' ~dt-- B dlc' (8_101)
where p' is the excess pressure (above atmospheric), v is the velocity, in
the ^-direction, of the air at any point, and where p is the density and B
the bulk modulus. The physical meaning of these equations is clear almost
without further discussion of the motion of gases. Both p' and v satisfy
wave equations, easily derived from Eqs. (8-101) :
aV l ay d\ l a 2 y
dx 2 c 2 at 2 ' ex 2 c 2 at 2
(8-102)
where
(b\ 112
-(!) '
(8-103)
and the power transmitted in the codirection per unit area is
P = p'v.
(8-104)
In the case of a plane electromagnetic wave traveling in the x-direction
and linearly polarized in the ^-direction, the analogous equations can be
shown to be (gaussian units)
dB z 6E V dE y dB z fo_ii\K\
-T7- = —C — ) -77" = — c — — > (8-105)
dt dx dt dx
where E u and B z are the y- and z-components of electric and magnetic field
intensities, and c is the speed of light. The components E v and B z satisfy
wave equations with wave velocity c, and the power transmitted in the
a;-direction per unit area is
As a final example, on a two-wire electrical transmission line, the volt-
age E across the line and the current i through the line satisfy the equa-
tions
dE _ 1 di di _ 1 bE ,„„„
'di-~cYx , di--Ldx-' (8_107)
where C is the shunt capacitance per unit length, and L is the series in-
ductance per unit length. Again we can derive wave equations for i and E
8-6] KINEMATICS OF MOVING FLUIDS 313
with the wave velocity
- (*r-
(8-108)
and again the power transmitted in the z-direction is
P = Ei. (8-109)
Thus the study of wave propagation in a string is applicable to a wide
variety of physical problems, many of them of greater practical and theo-
retical importance than the string itself. In many cases, our discussion
of the string as made up of a number of discrete particles is also of interest.
The electrical transmission line, for example, can be considered a limiting
case of a series of low-pass filters. An electrical network made up of series
inductances and shunt capacitances can be described by a set of equations
of the same form as our Eqs. (8-61), with analogous results. In the case
of sound waves, we are led by analogy to expect that at very high fre-
quencies, when the wavelength becomes comparable to the distance be-
tween molecules, the wave velocity will begin to depend on the frequency,
and that therefwill be a limiting frequency above which no wave propaga-
tion is possible.
8-6 Kinematics of moving fluids. In this section we shall develop the
kinematic concepts useful in studying the motion of continuously distrib-
uted matter, with particular reference to moving fluids. One way of
describing the motion of a fluid would be to attempt to follow the motion
of each individual point in the fluid, by assigning coordinates x, y, z to each
fluid particle and specifying these as functions of the time. We may, for ex-
ample, specify a given fluid particle by its coordinates, x , y , z , at an ini-
tial instant t = t . We can then describe the motion of the fluid by means
of functions x(x , y , Zo, 0. y( x o, Vo, *o, t), z(x , y 0> z , t) which determine
the coordinates x, y, z at time t of the fluid particle which was at x , y , z Q
at time t - This would be an immediate generalization of the concepts of
particle mechanics, and of the preceding treatment of the vibrating string.
This program originally due to Euler leads to the so-called "Lagrangian
equations" of fluid mechanics. A more convenient treatment for many
purposes, due also to Euler, is to abandon the attempt to specify the his-
tory of each fluid particle, and to specify instead the density and velocity
of the fluid at each point in space at each instant of time. This is the
method which we shall follow here. It leads to the "Eulerian equations"
of fluid mechanics. We describe the motion of the fluid by specifying the
density p(x, y, z, t) and the vector velocity v(x, y, z, t), at the point x, y, z
at the time t. We thus focus our attention on what is happening at a par-
ticular point in space at a particular time, rather than on what is happening
to a particular fluid particle.
314 THE MECHANICS OP CONTINUOUS MEDIA [CHAP. 8
Any quantity which is used in describing the state of the fluid, for ex-
ample the pressure p, will be a function [p{x, y, z, t)] of the space coordi-
nates x, y, z and of the time t; that is, it will have a definite value at each
point in space and at each instant of time. Although the mode of descrip-
tion we have adopted focuses attention on a point in space rather than on
a fluid particle, we shall not be able to avoid following the fluid particles
themselves, at least for short time intervals dt. For it is to the particles,
and not to the space points, that the laws of mechanics apply. We shall be
interested, therefore, in two time rates of change for any quantity, say p.
The rate at which the pressure is changing with time at a fixed point in
space will be the partial derivative with respect to time (dp/dt) ; it is itself
a function of x, y, z, and t. The rate at which the pressure is changing with
respect to a point moving along with the fluid will be the total derivative
dv^djp^dpdxdpdydpdz
dt dt ~ r dx dt^~ dy dt ^ dz dt' {0 V)
where dx/dt, dy/dt, dz/dt are the components of the fluid velocity v. The
change in pressure, dp, occurring during a time dt, at the position of a
moving fluid particle which moves from x, y, z to x -\- dx, y + dy, z -\- dz
during this time, will be
dp — p(x + dx,y + dy, z + dz, t + dt) — p(x, y, z, t)
and if dt — * 0, this leads to Eq. (8-110). We can also write Eq. (8-110) in
the forms:
dt ~ dt +Vx dx + Vy dy + V *dz
= £ + *.£ + *£ + '.£ (8-Hl)
and
1 = 1 + '-". <nw>
where the second expression is a shorthand for the first, in accordance with
the conventions for using the symbol V. The total derivative dp/dt is
also a function of x, y, z, and t. A similar relation holds between partial
and total derivatives of any quantity, and we may write, symbolically,
l=4+ v - v ' < 8 - ii3 >
where total and partial derivatives have the meaning defined above.
Let us consider now a small volume dV of fluid, and we shall agree that
dV always designates a volume element which moves with the fluid, so
8-6]
KINEMATICS OF MOVING FLUIDS
315
Fig. 8-4. A moving, expanding element of fluid.
that it always contains the same fluid particles. In general, the volume
SV will then change with time, and we wish to calculate this rate of change.
Let us assume that SV is in the form of a rectangular box of dimensions
Sx, Sy, Sz (Fig. 8-4) :
SV = Sx Sy Sz. (8-114)
The a;-component of fluid velocity v x may be different at the left and right
faces of the box. If so, Sx will change with time at a rate equal to the
difference between these two velocities :
and, similarly,
dt
Sx
dv x
dx
Sx,
d „ dVy „
d . dv z
-77 Sz = — Sz.
dt dz
(8-115)
The time rate of change of SV is then
— SV = Sy Sz 37 Sx + Sx Sz^-Sy+ Sx Sy -j: Sz
dt dt dt dt
= (t + ^ + ?^*^
dy
dz
)
and finally,
dt
SV = V-v 57.
(8-116)
This derivation is not very rigorous, but it gives an insight into the
meaning of the divergence V-v. The derivation can be made rigorous by
keeping careful track of quantities that were neglected here, like the de-
316 THE MECHANICS OF CONTIXUOCS MEDIA [CHAP. 8
pendence of v x upon y and 2, and showing that we arrive at Eq. (8-1 1G)
in the limit as SV — » 0. However, there is an easier way to give a more
rigorous proof of Eq. (8-116). Let us consider a volume V of fluid which
is composed of a number of elements 8V:
V = E SV. (8-117)
If we sum the left side of Eq. (8-116), we have
Ea**- »£"-£• (s-118)
The summation sign lie re really represents an integration, since we mean
to pass to the limit SV — ► 0, but the algebraic steps in Eq. (8-118) would
look rather unfamiliar if the integral sign were used. Now let us sum the
right side of Eq. (8-116), this time passing to the limit and using the in-
tegral sign, in order that we may apply Gauss' divergence theorem [Eq.
(3-115)]:
£ Vt«F= ff fv-vdV
= I Jn*v dS, (8-119)
where & is the surface bounding the volume V, and n is the outward
normal unit vector. Since n-v is the outward component of velocity
of the surface element dS, the volume added to V by the motion of dS in a
time dt will be n-v dt dS (Fig. 8-5), and hence the last line in Eq. (8-119)
is the proper expression for the rate of increase in volume :
dV r r
—■ = j Jn-vdS. (8-120)
s
Therefore Eq. (8-116) must be the correct expression for the rate of
Fig. 8-5. Increase of volume due to motion of surface.
8-6] KINEMATICS OP MOVING FLUIDS 317
increase of a volume element, since it gives the correct expression for the
rate of increase of any volume V when summed over V. Note that the
proof is independent of the shape of SV. We have incidentally derived an
expression for the time rate of change of a volume V of moving fluid:
= fffV'VdV. (8-121)
dV
dt
v
If the fluid is incompressible, then the volume of every element of fluid
must remain constant :
j t &V = 0, (8-122)
and consequently, by Eq. (8-116),
Vv = 0. (8-123)
No fluid is absolutely incompressible, but for many purposes liquids may
be regarded as practically so and, as we shall see, even the compressibility
of gases may often be neglected.
Now the mass of an element of fluid is
$m = p8V, (8-124)
and this will remain constant even though the volume and density may
not:
|a» = |<pl7) = 0. (8-125)
Let us carry out the differentiation, making use of Eq. (8-116) :
or, when SV is divided out,
^ + p vv = 0. (8-126)
By utilizing Eq. (8-113), we can rewrite this in terms of the partial deriva-
tives referred to a fixed point in space:
-£ + v- Vp + pV-v = 0.
318 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
The last two terms can be combined, using the properties of V as a symbol
of differentiation:
g+V(pv) = 0. (8-127)
This is the equation of continuity for the motion of continuous matter. It
states essentially that matter is nowhere created or destroyed; the mass
8m in any volume SV moving with the fluid remains constant.
We shall make frequent use in the remainder of this chapter of the properties
of the symbol V, which were described briefly in Section 3-6. The operator V
has the algebraic properties of a vector and, in addition, when a product is in-
volved, it behaves like a differentiation symbol. The simplest way to perform
this sort of manipulation, when V operates on a product, is first to write a sum
of products in each of which only one factor is to be differentiated. The factor
to be differentiated may be indicated by underlining it. Then each term may be
manipulated according to the rules of vector algebra, except that the underlined
factor must be kept behind the V symbol. When the underlined factor is the only
one behind the V symbol, or when all other factors are separated out by paren-
theses, the underline may be omitted, as there is no ambiguity as to what factor
is to be differentiated by the components of V. As an example, the relation
between Eqs. (8-126) and (8-127) is made clear by the following computation:
V-(pv) = V-(pv) + V-(pv)
= (Vp)«v+ pV-v
= (Vp)-v+ pV-v
= vVp+pV-v. (8-128)
Any formulas arrived at in this way can always be verified by writing out both
sides in terms of components, and the reader should do this a few times to con-
vince himself. However, it is usually far less work to make use of the properties
of the V symbol.
We now wish to calculate the rate of flow of mass through a surface S
fixed in space. Let dS be an element of surface, and let n be a unit vector
normal to dS. If we construct a cylinder by moving dS through a distance
v dt in the direction of — v, then in a time dt all the matter in this cylinder
will pass through the surface dS (Fig. 8-6). The amount of mass in this
cylinder is
pn-v dt dS,
where n-v dt is the altitude perpendicular to the face dS. The rate of
flow of mass through a surface S is therefore
dm
dt
ffpn-vdS = JJn- (pv) dS. (8-129)
8-6]
KINEMATICS OF MOVING FLUIDS
319
Fig. 8-6. Flow of fluid through a surface element.
If n-v is positive, the mass flow across S is in the direction of n; if n-v
is negative, the mass flow is in the reverse direction. We see that pv, the
momentum density, is also the mass current, in the sense that its com-
ponent in any direction gives the rate of mass flow per unit area in that
direction. We can now give a further interpretation of Eq. (8-127) by
integrating it over a fixed volume V bounded by a surface S with outward
normal n:
V V
(8-130)
Since the volume V here is a fixed volume, we can take the time differenti-
ation outside the integral in the first term. If we apply Gauss' divergence
theorem to the second integral, we can rewrite this equation:
dt
JffpdV= -Jfn-(pv)dS.
(8-131)
This equation states that the rate of increase of mass inside the fixed vol-
ume V is equal to the negative of the rate of flow of mass outward across
the surface. This result emphasizes the physical interpretation of each
term in Eq. (8-127). In particular, the second term evidently represents
the rate of flow of mass away from any point. Conversely, by starting
with the self-evident equation (8-131) and working backwards, we have
an independent derivation of Eq. (8-127).
Equations analogous to Eqs. (8-126), (8-127), (8-129), and (8-131)
apply to the density, velocity, and rate of flow of any physical quantity.
An equation of the form (8-127) applies, for example, to the flow of electric
charge, if p is the charge density and pv the electric current density.
320
THE MECHANICS OF CONTINUOUS MEDIA
[CHAP. 8
/ 'i
(a)
(b)
Fig. 8-7. Meaning of nonzero curl v. (a) A vortex, (b) A transverse velocity
gradient.
The curl of the velocity V X v is a concept which is useful in describing
fluid flow. To understand its meaning, we compute the integral of the
normal component of curl v across a surface S bounded by a curve C. By
Stokes' theorem (3-117), this is
J fn'(V X v)dS = f v-dr,
(8-132)
where the line integral is taken around C in the positive sense relative to
the normal n, as previously defined. If the curve C surrounds a vortex in
the fluid, so that v is parallel to dr around C (Fig. 8-7), then the line in-
tegral on the right is positive and measures, in a sense, the rate at which
the fluid is whirling around the vortex. Thus V X v is a sort of measure
of the rate of rotation of the fluid per unit area; hence the name curl v.
Curl v has a nonzero value in the neighborhood of a vortex in the fluid.
Curl v may also be nonzero, however, in regions where there is no vortex,
that is, where the fluid does not actually circle a point, provided there is
a transverse velocity gradient. Figure 8-7 illustrates the two cases. In
each case, the line integral of v counterclockwise around the circle C will
have a positive value. If the curl of v is zero everywhere in a moving fluid,
the flow is said to be irrotational. Irrotational flow is important chiefly
because it presents fairly simple mathematical problems. If at any point
V X v = 0, then an element of fluid at that point will have no net
angular velocity about that point, although its shape and size may be
changing.
We arrive at a more precise meaning of curl v by introducing a co-
ordinate system rotating with angular velocity u. If v' designates the
velocity of the fluid relative to the rotating system, then by Eq. (7-33),
v = v' + to X r,
where r is a vector from the axis of rotation (whose location does not mat-
ter in this discussion) to a point in the fluid. Curl v is now
8-7] EQUATIONS OF MOTION FOR AN IDEAL FLUID 321
Vxv=vxv' + Tx(«xr)
= V X v' 4- wV-r — wT r
= V X v' + 3« — <o
= V X v' + 2«,
where the second line follows from Eq. (3-35) for the triple cross product,
and the third line by direct calculation of the components in the second
and third terms. If we set
w = a V X v, (8-133)
then
V X v' = 0. (8-134)
Thus if V X v j£ at a point P, then in a coordinate system rotating with
angular velocity « = |V X y, the fluid flow is irrotational at the point P.
We may therefore interpret |V X v as the angular velocity of the fluid
near any point. If V X v is constant, then it is possible to introduce a
rotating coordinate system in which the flow is irrotational everywhere.
8-7 Equations of motion for an ideal fluid. For the remainder of this
chapter, except in the last section, we shall consider the motion of an ideal
fluid, that is, one in which there are no shearing stresses, even when the
fluid is in motion. The stress within an ideal fluid consists in a pressure p
alone. This is a much greater restriction in the case of moving fluids than
in the case of fluids in equilibrium (Section 5-11). A fluid, by definition,
supports no shearing stress when in equilibrium, but all fluids have some
viscosity and therefore there are always some shearing stresses between
layers of fluid in relative motion. An ideal fluid would have no viscosity,
and our results for ideal fluids will therefore apply only when the viscosity
is negligible.
Let us suppose that, in addition to the pressure, the fluid is acted on
by a body force of density f per unit volume, so that the body force acting
on a volume element SV of fluid is f 8V. We need, then, to calculate the
force density due to pressure. Let us consider a volume element SV =
Sx 8y 8z in the form of a rectangular box (Fig. 8-8). The force due to
pressure on the left face of the box is p 8y Sz, and acts in the ^-direction.
The force due to pressure on the right face of the box is also p by 8z, and
acts in the opposite direction. Hence the net a>component of force 8F X
on the box depends upon the difference in pressure between the left and
right faces of the box:
8F X = ( - |£ Sx) Sy 8z. (8-135)
322
THE MECHANICS OF CONTINUOUS MEDIA
[CHAP. 8
Sx
Fig. 8-8. Force on a volume element due to pressure.
A similar expression may be derived for the components of force in the
y- and z-directions. The total force on the fluid in the box due to pressure
is then
8F
-(-
.dp _ .dp
dx dy
SV
= — Vp SV.
(8-136)
The force density per unit volume due to pressure is therefore — Vp.
This result was also obtained in Section 5-11 [Eq. (5-172)].
We can now write the equation of motion for a volume element SV
of fluid:
p5y^= f 57 - VpSV.
(8-137)
This equation is usually written in the form
p-+Vp = f.
(8-138)
By making use of the relation (8-113), we may rewrite this in terms of
derivatives at a fixed point :
+ V- W -\ Vp = - :
P P
at
(8-139)
where f/p is the body force per unit mass. This is Euler's equation of
motion for a moving fluid.
If the density p depends only on the pressure p, we shall call the fluid
homogeneous. This definition does not imply that the density is uniform.
An incompressible fluid is homogeneous if its density is uniform. A com-
8-8] CONSERVATION LAWS FOR FLUID MOTION
323
pressible fluid of uniform chemical composition and uniform temperature
throughout is homogeneous. When a fluid expands or contracts under the
influence of pressure changes, work is done by or on the fluid, and part of
this work may appear in the form of heat. If the changes in density occur
sufficiently slowly so that there is adequate time for heat flow to maintain
the temperature uniform throughout the fluid, the fluid may be considered
homogeneous within the meaning of our definition. The relation between
density and pressure is then determined by the equation of state of the
fluid or by its isothermal bulk modulus (Section 5-11). In some cases,
changes in density occur so rapidly that there is no time for any appreciable
flow of heat. In such cases the fluid may also be considered homogeneous,
and the adiabatic relation between density and pressure or the adiabatic
bulk modulus should be used. In cases between these two extremes, the
density will depend not only on pressure, but also on temperature, which,
in turn, depends upon the rate of heat flow between parts of the fluid at
different temperatures.
In a homogeneous fluid, there are four unknown functions to be de-
termined at each point in space and time, the three components of velocity
v, and the pressure p. We have, correspondingly, four differential equations
to solve, the three components of the vector equation of motion (8-139),
and the equation of continuity (8-127). The only other quantities ap-
pearing in these equations are the body force, which is assumed to be
given, and the density p, which can be expressed as a function of the
pressure. Of course, Eqs. (8-139) and (8-127) have a tremendous variety
of solutions. In a specific problem we would need to know the conditions
at the boundary of the region in which the fluid is moving and the values
of the functions v and p at some initial instant. In the following sections,
we shall confine our attention to homogeneous fluids. In the intermediate
case mentioned at the end of the last paragraph, where the fluid is in-
homogeneous and the density depends on both pressure and temperature,
we have an additional unknown function, the temperature, and we will
need an additional equation determined by the law of heat flow. We shall
not consider this case, although it is a very important one in many
problems.
8-8 Conservation laws for fluid motion. Inasmuch as the laws of fluid
motion are derived from Newton's laws of motion, we may expect that
appropriate generalizations of the conservation laws of momentum, energy,
and angular momentum also hold for fluid motion. We have already had
an example of a conservation law for fluid motion, namely, the equation
of continuity [Eq. (8-127) or (8-131)], which expresses the law of con-
servation of mass. Mass is conserved also in particle mechanics, but we
did not find it necessary to write an equation expressing this fact.
324 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
A conservation law in fluid mechanics may be written in many equiva-
lent forms. It will be instructive to study some of these in order to get a
clearer idea of the physical meaning of the various mathematical expres-
sions involved. Let p be the density of any physical quantity: mass,
momentum, energy, or angular momentum. Then the simplest form of
the conservation law for this quantity will be equation (8-125), which
states that the amount of this quantity in an element SV of fluid remains
constant. If the quantity in question is being produced at a rate Q per
unit volume, then Eq. (8-125) should be generalized:
1 (p SV) = Q SV. (8-140)
This is often called a conservation law for the quantity p. It states that
this quantity is appearing in the fluid at a rate Q per unit volume, or disap-
pearing if Q is negative. In the sense in which we have used the term in
Chapter 4, this should not be called a conservation law except when
Q = 0. By a derivation exactly like that which led to Eq. (8-127), we
can rewrite Eq. (8-140) as a partial differential equation:
f+V-(pv) = Q. (8-141)
at
This is probably the most useful form of conservation law. The meaning
of the terms in Eq. (8-141) is brought out by integrating each term over a
fixed volume V and using Gauss' theorem,* as in the derivation of Eq.
(8-131) :
^ t fffpdV + ffn-vpdS = IJJQdV. (8-142)
According to the discussion preceding Eq. (8-129), this equation states
that the rate of increase of the quantity within V, plus the rate of flow
outward across the boundary S, equals the rate of appearance due to
sources within V. Another form of the conservation law which is some-
times useful is obtained by summing equation (8-140) over a volume V
moving with the fluid :
* If p is a vector, as in the case of linear or angular momentum density, then a
generalized form of Gauss' theorem [mentioned in Section 5-11 in connection
with Eq. (5-178)] must be used.
8 ~ 8 J CONSERVATION LAWS FOR FLUID MOTION 325
If we pass to the limit SV -> 0, the summations become integrations:
ifIfpdV = JffQdV. (8-144)
v v
The surface integral which appears in the left member of Eq. (8-142)
does not appear in Eq. (8-144) ; since the volume V moves with the fluid,
there is no flow across its boundary. Since Eqs. (8-140), (8-141), (8-142),
and (8-144) are all equivalent, it is sufficient to derive a conservation law
in any one of these forms. The others then follow. Usually it is easiest to
derive an equation of the form (8-140), starting with the equation of
motion in the form (8-138). We can also start with Eq. (8-139) and de-
rive a conservation equation in the form (8-141), but a bit more manipula-
tion is usually required.
In order to derive a conservation law for linear momentum, we first
note that the momentum in a volume element SV is pv SV. The mo-
mentum density per unit volume is therefore pv, and this quantity will
play the role played by p in the discussion of the preceding paragraph. In
order to obtain an equation analogous to Eq. (8-140), we start with the
equation of motion in the form (8-138), which refers to a point moving
with the fluid, and multiply through by the volume 57 of a small fluid
element :
dv
p SV It + Vp SV = f SV - (8-145)
Since p SV = Sm is constant, we may include it in the time derivative:
- (pv SV) = (f - Vp) SV. (8-146)
The momentum of a fluid element, unlike its mass, is not, in general, con-
stant. This equation states that the time rate of change of momentum
of a moving fluid element is equal to the body force plus the force due to
pressure acting upon it. The quantity f — Vp here plays the role of Q
in the preceding general discussion. Equation (8-146) can be rewritten
in any of the forms (8-141), (8-142), and (8-144). For example, we may
write it in the form (8-144) :
d
dt
] I ' JpvdV = ffffdV - fffvpdV. (8-147)
We can now apply the generalized form of Gauss' theorem [Eq. (5-178)]
to the second term on the right, to obtain
326 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
| fffpv dV = ffft dV + ff-np dS, (8-148)
where S is the surface bounding V.
This equation states that the time rate of change of the total linear
momentum in a volume V of moving fluid is equal to the total external
force acting on it. This result is an immediate generalization of the linear
momentum theorem (4-7) for a system of particles. The internal forces,
in the case of a fluid, are represented by the pressure within the fluid. By
the application of Gauss' theorem, we have eliminated the pressure within
the volume V, leaving only the external pressure across the surface of V.
It may be asked how we have managed to eliminate the internal forces
without making explicit use of Newton's third law, since Eq. (8-138),
from which we started, is an expression only of Newton's first two laws.
The answer is that the concept of pressure itself contains Newton's third
law implicitly, since the force due to pressure exerted from left to right
across any surface element is equal and opposite to the force exerted from
right to left across the same surface element. Furthermore, the points of
application of these two forces are the same, namely, at the surface ele-
ment. Both forces necessarily have the same line of action, and there is no
distinction between the weak and strong forms of Newton's third law.
The internal pressures will therefore also be expected to cancel out in the
equation for the time rate of change of angular momentum. A similar
remark applies to the forces due to any kind of stresses in a fluid or a solid;
Newton's third law in strong form is implicitly contained in the concept of
stress.
Equations representing the conservation of angular momentum, analo-
gous term by term with Eqs. (8-140) through (8-144), can be derived by
taking the cross product of the vector r with either Eq. (8-138) or (8-139),
and suitably manipulating the terms. The vector r is here the vector from
the origin about which moments are to be computed to any point in the
moving fluid or in space. This development is left as an exercise. The
law of conservation of angular momentum is responsible for the vortices
formed when a liquid flows out through a small hole in the bottom of a
tank. The only body force here is gravity, which exerts no torque about
the hole, and it can be shown that if the pressure is constant, or depends
only on vertical depth, there is no net vertical component of torque across
any closed surface due to pressure. Therefore the angular momentum of
any part of the fluid remains constant. If a fluid element has any angular
momentum at all initially, when it is some distance from the hole, its angu-
lar velocity will have to increase in inverse proportion to the square of its
distance from the hole in order for its angular momentum to remain con-
stant as it approaches the hole.
8-8] CONSERVATION LAWS FOR FLUID MOTION 327
In order to derive a conservation equation for the energy, we take the
dot product of v with Eq. (8-146), to obtain
j t (ipv 2 SV) = v(f - Vp) SV. (8-149)
This is the energy theorem in the form (8-140). In place of the density p,
we have here the kinetic energy density \pv 2 . The rate of production of
kinetic energy per unit volume is
Q = v (f - Vp). (8-150)
In analogy with our procedure in particle mechanics, we shall now try
to define additional forms of energy so as to include as much as possible
of the right member of Eq. (8-149) under the time derivative on the left.
We can see how to rewrite the second term on the right by making use of
Eqs. (8-113) and (8-116):
l(p d V) = ^SV + p^^
so that
dv
= m 6V + v-Vp87 + pV-vSV, (8-151)
-v Vp SV = - ~ (p 57) + & SV + pV-v SV. (8-152)
Let us now assume that the body force f is a gravitational force:
f = Pg = pvg, (8-153)
where g is the gravitational potential [Eq. (6-16)], i.e., the negative poten-
tial energy per unit mass due to gravitation. The first term on the right
in Eq. (8-149) is then
vf SV = (v vg)p SV = (^ - g )p SV
= !<P9«F)-pf£-a7, (8-154)
since p SV = Sm is constant. With the help of Eqs. (8-152) and (8-154),
Eq. (8-149) can be rewritten :
dt
; [(*Pf 2 + V ~ P9) SV) = (^ - pg) SV + pv-v SV. (8-155)
The pressure p here plays the role of a potential energy density whose
328 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
negative gradient gives the force density due to pressure [Eq. (8-136)].
The time rate of change of kinetic energy plus gravitational potential
energy plus potential energy due to pressure is equal to the expression on
the right.
Ordinarily, the gravitational field at a fixed point in space will not
change with time (except perhaps in applications to motions of gas clouds
in astronomical problems). If the pressure at a given point in space is
constant also, then the first term on the right vanishes. What is the sig-
nificance of the second term? For an incompressible fluid, V-v = 0,
and the second term would vanish also. We therefore suspect that it
represents energy associated with compression and expansion of the fluid
element SV. Let us check this hypothesis by calculating the work done in
changing the volume of the element SV. The work dW done by the fluid
element SV, through the pressure which it exerts on the surrounding fluid
when it expands by an amount d SV, is
dW = pd SV. (8-156)
The rate at which energy is supplied by the expansion of the fluid element
is, by Eq. (8-116),
dW d SV _ _ T7 . / 1C » N
-dT = p -dT = pV ' v8V ' (8 ~ 157)
which is just the last term in Eq. (8-155). So far, all our conservation
equations are valid for any problem involving ideal fluids. If we restrict
ourselves to homogeneous fluids, that is, fluids whose density depends only
on the pressure, we can define a potential energy associated with the ex-
pansion and contraction of the fluid element SV. We shall define the
potential energy u 8m on the fluid element SV as the negative work done
through its pressure on the surrounding fluid when the pressure changes
from a standard pressure p to any pressure p. The potential energy per
unit mass u will then be a function of p :
-f
JPo
uSm= - / pd SV. (8-158)
J Pa
The volume SV = Sm/p is a function of pressure, and we may rewrite this
in various forms:
J Pa /
<Pa P*
dp
I
I
^f-dp (8-159)
po P 2 d V
V
PaP B
8-9] STEADY FLOW 329
where the last step makes use of the definition of the bulk modulus [Eq.
(5-116)]. The time rate of change of u is, by Eqs. (8-158) or (8-159)
and (8-116),
d{u 8m) d 8V
~dT~ = ~ p -dT= -p v ' v sv - ( 8 -!60)
We can now include the last term on the right in Eq. (8-155) under the
time derivative on the left :
j t l(W + p-ps + pu) SV] = (& _ p g?) hV . (8-161)
The interpretation of this equation is clear from the preceding discussion.
It can be rewritten in any of the forms (8-141), (8-142), and (8-144).
If p and § are constant at any fixed point in space, then the total kinetic
plus potential energy of a fluid element remains constant as it moves
along. It is convenient to divide by 5m = p 57 in order to eliminate refer-
ence to the volume element :
This is Bernoulli's theorem. The term dQ/dt is practically always zero;
we have kept it merely to make clear the meaning of the term (l/p)(dp/dt),
which plays a similar role and is not always zero. When both terms on the
right are zero, as in the case of steady flow, we have, for a point moving
along with the fluid,
y2 , V
2" + - — 9 + w=a constant. (8-163)
Other things being equal, that is if u, g, and p are constant, the pressure of
a moving fluid decreases as the velocity increases. For an incompressible
fluid, p and u are necessarily constant.
The conservation laws of linear and angular momentum apply not only
to ideal fluids, but also, when suitably formulated, to viscous fluids and
even to solids, in view of the remarks made above regarding Newton's
third law and the concept of stress. The law of conservation of energy
(8-162) will not apply, however, to viscous fluids, since the viscosity is
due to an internal friction which results in a loss of kinetic and potential
energies, unless conversion of mechanical to heat energy by viscous
friction is included in the law. [Equation (8-155) applies in any case.]
8-9 Steady flow. By steady flow of a fluid we mean a motion of the
fluid in which all quantities associated with the fluid, velocity, density,
pressure, force density, etc., are constant in time at any given point in
330 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
space. For steady flow, all partial derivatives with respect to time can be
set equal to zero. The total time derivative, which designates the time
rate of change of a quantity relative to a point moving with the fluid, will
not in general be zero, but, by Eq. (8-113), will be
^ = v-V. (8-164)
at
The path traced out by any fluid element as it moves along is called a
streamline. A streamline is a line which is parallel at each point (x, y, z)
to the velocity v(x, y, z) at that point. The entire space within which the
fluid is flowing can be filled with streamlines such that through each point
there passes one and only one streamline. If we introduce along any
streamline a coordinate s which represents the distance measured along
the streamline from any fixed point, we can regard any quantity associated
with the fluid as a function of s along the streamline. The component of
the symbol V along the streamline at any point is d/ds, as we see if we
choose a coordinate system whose z-axis is directed along the streamline
at that point. Equation (8-164) can therefore be rewritten:
it = 4s- (8 - l65)
This equation is also evident from the fact that v = ds/dt. For example,
Eq. (8-162), in the case of steady flow, can be written:
i(£+- P -s +u )=°- (8 - i66)
The quantity in parentheses is therefore constant along a streamline.
The equation of continuity (8-127) in the case of steady flow becomes
V-(pv) = 0. (8-167)
If we integrate this equation over a fixed volume V, and apply Gauss'
theorem, we have
dS = 0, (8-168)
J Jn-(pv)
where S is the closed surface bounding V. This equation simply states
that the total mass flowing out of any closed surface is zero.
If we consider all the streamlines which pass through any (open) sur-
face S, these streamlines form a tube, called a tube of flow (Fig. 8-9). The
walls of a tube of flow are everywhere parallel to the streamlines, so that
no fluid enters or leaves it. A surface S which is drawn everywhere per-
pendicular to the streamlines and through which passes each streamline in
8-9] STEADY FLOW 331
Fio. 8-9. A tube of flow.
a tube of flow, will be called a cross section of the tube. If we apply Eq.
(8-168) to the closed surface bounded by the walls of a tube of flow and
two cross sections St and S 2 , then since n is perpendicular to v over the
walls of the tube, and n is parallel or antiparallel to v over the cross sec-
tions, we have
Jfpv dS - J jpv dS = 0, (8-169)
or
J J pv dS = I = a constant, (8-170)
8
where S is any cross section along a given tube of flow. The constant I
ia called the fluid current through the tube.
The energy conservation equation (8-161), when rewritten in the form
(8-141), becomes, in the case of steady flow,
V-[(W + p - pg + pu)v] = 0. (8-171)
This equation lias the same form as Eq. (8-167), and we can conclude in
the same way that the energy current is the same through any cross sec-
tion S of a tube of flow:
/ ' Jdpv 2 + V ~ P£ + pv)» dS = a constant. (8-172)
5
This result is closely related to Eq. (8-166).
If the flow ia not only steady, but also irrotational, then
V X v = (8-173)
everywhere. This equation ia analogous in form to Eq. (3-189) for a conservative
force, and we can proceed as in Section 3-12 to show that if Eq. (8-173) holds, it
332 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
is possible to define a velocity potential function <j>(x, y, z) by the equation
0(r) = fv-dr, (8-174)
where r„ is any fixed point. The velocity at any point will then be
v = V<f>. (8-175)
Substituting this in Eq. (8-167), we have an equation to be solved for <t>:
V-(pV<rt = 0. (8-176)
In the cases usually studied, the fluid can be considered incompressible, and this
becomes
V 2 4> = 0. (8-177)
This equation is identical in form with Laplace's equation (6-35) for the gravita-
tional potential in empty space. Hence the techniques of potential theory may
be used to solve problems involving irrotational flow of an incompressible fluid.
8-10 Sound waves. Let us assume a fluid at rest with pressure p ,
density po, in equilibrium under the action of a body force f , constant in
time. Equation (8-139) then becomes
fv Po = ^- (8-178)
Po Po
We may note that this equation agrees with Eq. (5-172) deduced in Sec-
tion 5-11 for a fluid in equilibrium. Let us now suppose that the fluid is
subject to a small disturbance, so that the pressure and density at any
point become
V = Po + V', (8-179)
P = Po + p', (8-180)
where p'«p and p' <C p. We assume that the resulting velocity v and
its space and time derivatives are everywhere very small. If we substi-
tute Eqs. (8-179) and (8-180) in the equation of motion (8-139), and
neglect higher powers than the first of p', p', v and their derivatives, mak-
ing use of Eq. (8-178), we obtain
%=-&*• (8 " 181)
Making a similar substitution in Eq. (8-127), we obtain
22-= -poV-v - Wpo. (8-182)
8-10] SOUND WAVES 333
Let us assume that the equilibrium density po is uniform, or nearly so, so
that Vp is zero or very small, and the second term can be neglected.
The pressure increment p' and density increment p' are related by the
bulk modulus according to Eq. (5-183) :
p' p'
This equation may be used to eliminate either p' or p' from Eqs. (8-181)
and (8-182). Let us eliminate p' from Eq. (8-182) :
dp'
-~= -BV-v. (8-184)
Equations (8-181) and (8-184) are the fundamental differential equations
for sound waves. The analogy with the form (8-101) for one-dimensional
waves is apparent. Here again we have two quantities, p' and v, such that
the time derivative of either is proportional to the space derivatives of the
other. In fact, if v = iv x and if v x and p' are functions of x alone, then
Eqs. (8-181) and (8-184) reduce to Eqs. (8-101).
We may proceed, in analogy with the discussion in Section 8-5, to
eliminate either v or p' from these equations. In order to eliminate v,
we take the divergence of Eq. (8-181) and interchange the order of differ-
entiation, again assuming p nearly uniform :
JL(v.v) = -lvV. (8-185)
We now differentiate Eq. (8-184) with respect to t, and substitute from
Eq. (8-185):
V V - i d -X = 0, (8-186)
c 2 dt 2
where
Vpo/
(8-187)
This is the three-dimensional wave equation, as we shall show presently.
Formula (8-187) for the speed of sound waves was first derived by Isaac
Newton, and applies either to liquids or gases. For gases, Newton as-
sumed that the isothermal bulk modulus B = p should be used, but Eq.
(8-187) does not then agree with the experimental values for the speed of
sound. The sound vibrations are so rapid that they should be treated as
adiabatic, and the adiabatic bulk modulus B = Jp should be used, where
7 is the ratio of specific heat at constant pressure to that at constant
334 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
volume.* Formula (8-187) then agrees with the experimental values of c.
If we eliminate p' by a similar process, we obtain a wave equation for v:
^ - ±§ = 0. (8-188)
In deriving Eq. (8-188), it is necessary to use the fact that V X v = 0. It
follows from Eq. (8-181) that V X v is in any case independent of time, so
that the time-dependent part of v which is present in a sound wave is ir-
rotational. [We could add to the sound wave a small steady flow with
V X v ^ 0, without violating Eqs. (8-181) and (8-182).]
In order to show that Eq. (8-186) leads to sound waves traveling with
speed c, we note first that if p' is a function of x and t alone, Eq. (8-186)
becomes
This is of the same form as the one-dimensional wave equation (8-6), and
therefore has solutions of the form
p' = f(x — ct). (8-190)
This is called a plane wave, for at any time t the phase x — ct and the pres-
sure p' are constant along any plane (x = a constant) parallel to the yz-
plane. A plane wave traveling in the direction of the unit vector n will
be given by
p' = /(n-r - ct), (8-191)
where r is the position vector from the origin to any point in space. To
see that this is a wave in the direction n, we rotate the coordinate system
until the z-axis lies in this direction, in which case Eq. (8-191) reduces
to Eq. (8-190). The planes / = a constant, at any time t, are now per-
pendicular to n, and travel in the direction of n with velocity c. We can
see from the argument just given that the solution (8-191) must satisfy
Eq. (8-186), or we may verify this by direct computation, for any coor-
dinate system:
v r' = % v * = %*' < 8 " 192 >
where
$ = n-r - ct, (8-193)
* Millikan, Roller, and Watson, Mechanics, Molecular Physics, Heat, and
Sound. Boston: Ginn and Co., 1937. (Pages 157, 276.)
8-10] SOUND WAVES 335
and, similarly,
W-g».V« = g..n = g, (8-194)
dp dp W c dp (8 195)
so that Eq. (8-186) is satisfied, no matter what the function /(£) may be.
Equation (8-188) will also have plane wave solutions:
v = h(n'.r - ct), (8-196)
corresponding to waves traveling in the direction n' with velocity c, where
h is a vector function of £' = n'-r — ct. To any given pressure wave of
the form (8-191) will correspond a velocity wave of the form (8-196), re-
lated to it by Eqs. (8-181) and (8-182). If we calculate dv/dt from Eq.
(8-196), and Vp' from Eq. (8-191), and substitute in Eq. (8-181), we will
have
— — n it
de ~ (Bpo) 1/2 d*' (8-197)
Equation (8-197) must hold at all points r at all times t. The right mem-
ber of this equation is a function of £ and is constant for a constant £.
Consequently, the left member must be constant when £ is constant, and
must be a function only of £, which implies that f ' = { (or at least that
£' is a function of £), and hence n' = n. This is obvious physically, that
the velocity wave must travel in the same direction as the pressure wave.
We can now set £' = £, and solve Eq. (8-197) for h:
h = (Sk 1 **' < 8 ~ 198 >
where the additive constant is zero, since both p' and v are zero in a region
where there is no disturbance. Equations (8-198), (8-196), and (8-190)
imply that for a plane sound wave traveling in the direction n, the pressure
increment and velocity are related by the equation
V= (5^ n ' < 8 - 199 >
where v, of course, is here the velocity of a fluid particle, not that of the
wave, which is en. The velocity of the fluid particles is along the direction
of propagation of the sound wave, so that sound waves in a fluid are longi-
tudinal. This is a consequence of the fact that the fluid will not support
a shearing stress, and is not true of sound waves in a solid, which may be
either longitudinal or transverse.
336 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
A plane wave oscillating harmonically in time with angular frequency w
may be written in the form
p' = A cos (k-r - cot) = Re Ae*^ 1 -"* , (8-200)
where k, the wave vector, is given by
k = -n. (8-201)
If we consider a surface perpendicular to n which moves back and
forth with the fluid as the wave goes by, the work done by the pressure
across this surface in the direction of the pressure is, per unit area per unit
time,
P = pv. (8-202)
If v oscillates with average value zero, then since p = p + p', where p is
constant, the average power is
Pav = {p'vU = ~^^2 ' (8-203)
where we have made use of Eq. (8-199). This gives the amount of energy
per unit area per second traveling in the direction n.
The three-dimensional wave equation (8-186) has many other solutions
corresponding to waves of various forms whose wave fronts (surfaces of
constant phase) are of various shapes, and traveling in various directions.
As an example, we consider a spherical wave traveling out from the origin.
The rate of energy flow is proportional to p' 2 (a small portion of a spherical
wave may be considered plane), and we expect that the energy flow per
unit area must fall off inversely as the square of the distance, by the
energy conservation law. Therefore p' should be inversely proportional to
the distance r from the origin. We are hence led to try a wave of the form
P' = ^f(r-ct). (8-204)
This will represent a wave of arbitrary time-dependence, whose wave
fronts, £ = r — ct = a constant, are spheres expanding with the velocity
c. It can readily be verified by direct computation, using either rectangular
coordinates, or using spherical coordinates with the help of Eq. (3-124),
that the solution (8-204) satisfies the wave equation (8-186).
A slight difficulty is encountered with the above development if we attempt to
apply to a sound wave the expressions for energy flow and mass flow developed
in the two preceding sections. The rate of flow of mass per unit area per second,
8-11] NORMAL VIBRATIONS OF FLUID IN A RECTANGULAR BOX 337
by Eqs. (8-199), (8-180), and (8-183), is
pv = po(l + ^)^ T7 ,n.
We should expect that pv would be an oscillating quantity whose average value
is zero for a sound wave, since there should be no net flow of fluid. If we average
the above expression, we have
1/2
<pv)av = g^ «p' 2 )av + B(p% v )ll,
so that there is a small net flow of fluid in the direction of the wave, unless
(P')av=-^^- (8-205)
If Eq. (8-205) holds, so that there is no net flow of fluid, then it can be shown
that, to second-order terms in p' and v, the energy current density given by Eq.
(8-161) is, on the average, for a sound wave,
(dpv 2 + V - P9 + P«)v)av = { p QB l7/2 n > (8-206)
in agreement with Eq. (8-203). When approximations are made in the equa-
tions of motion, we may expect that the solutions will satisfy the conservation
laws only to the same degree of approximation. By adding second-order (or
higher) terms like (8-205) to a first-order solution, we can of course satisfy the
conservation laws to second-order terms (or higher).
8-11 Normal vibrations of fluid in a rectangular box. The problem of
the vibrations of a fluid confined within a rigid box is of interest not only
because of its applications to acoustical problems, but also because the
methods used can be applied to problems in electromagnetic vibrations,
vibrations of elastic solids, wave mechanics, and all phenomena in physics
which are described by wave equations. In this section, we consider a
fluid confined to a rectangular box of dimensions L x L y L z .
We proceed as in the solution of the one-dimensional wave equation in
Section 8-2. We first assume a solution of Eq. (8-186) of the form
V' = U(x,y,z)@(t). (8-207)
Substitution in Eq. (8-186) leads to the equation
Again we argue that since the left side depends only on x, y, and z, and the
338 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
right side only on t, both must be equal to a constant, which we shall call
-u 2 /c 2 :
d*e
+ « J © = 0, (8-209)
dP
2
V 2 C/+\[/ = 0. (8-210)
The solution of Eq. (8-209) can be written :
= A cos ut + B sin ut, (8-211)
or
© = Ae-*", (8-212)
where A and B are constant. The form (8-212) leads to traveling waves
of the form (8-200). We are concerned here with standing waves, and we
therefore choose the form (8-211). In order to solve Eq. (8-210), we again
use the method of separation of variables, and assume that
U(x, y, z) = X(x)Y(y)Z(z). (8-213)
Substitution in Eq. (8-210) leads to the equation
1 £X 1 d*Y 1 d*Z J
X dx 2 ** Y dyz ~*~ Z dz* c 2 " ^ 8 ;
This can hold for all x, y, z only if each term on the left is constant. We
shall call these constants — fc 2 ., — fc 2 , — fc 2 , so that
g + er=0, + fc 2 7 = O, g + fc 2 Z = 0, (8-215)
where 2
kl + fc 2 + fc 2 = ^ • (8-216)
The solutions of Eqs. (8-215) in which we are interested are
X = C x cos k x x + D x sin k x x,
Y = C y cos k y y + D y sin k v y, (8-217)
Z = C 2 cos k z z + D z sin k z z.
If we choose complex exponential solutions for X, Y, Z, and 0, we arrive at
the traveling wave solution (8-200), where k X) k v , k z are the components
of the wave vector k.
8-11] NORMAL VIBRATIONS OF FLUID IN A RECTANGULAR BOX 339
We must now determine the appropriate boundary conditions to be
applied at the walls of the box, which we shall take to be the six planes
x = 0, x = L x , y = 0, y = L v , z = 0, z = L z . The condition is evidently
that the component of velocity perpendicular to the wall must vanish at
the wall. At the wall x = 0, for example, v x must vanish. According to
Eq. (8-181),
dv x 1 dp'
af=-^ar <Mi8)
We substitute for p' from Eqs. (8-207), (8-211), (8-213), and (8-217):
dv k YZ
-JX = ~ — (A cos cot + B sin cot)(—C x sin k x x + D x cos k x x).
(8-219)
Integrating, we have
k YZ
v x = ~— (A sin tot — B cos ut)(—C x sin k x x + D x cos k x x)
(8-220)
plus a function of x, y, z, which vanishes, since we are looking for oscillating
solutions. In order to ensure that v x vanishes at x = 0, we must set
D x = 0, i.e., choose the cosine solution for X in Eq. (8-217). This means
that the pressure p' must oscillate at maximum amplitude at the wall.
This is perhaps obvious physically, and could have been used instead of the
condition v x = 0, which, however, seems more self-evident. The velocity
component perpendicular to a wall must have a node at the wall, and the
pressure must have an antinode. Similarly, the pressure must have an
antinode (maximum amplitude of oscillation) at the wall x — L x :
cos k x L x = ±1, (8-221)
so that
k x = 1 ^-, I = 0, 1, 2, . . . . (8-222)
By applying similar considerations to the four remaining walls, we con-
clude that D y — D z — 0, and
7 m7r r> t o
k y =-rj~, m = 0, 1, 2, . . . ,
Lly
(8-223)
fc z = j- , n = 0, 1, 2,
Li z
For each choice of three integers I, m, n, there is a normal mode of vibra-
340 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
tion of the fluid in the box. The frequencies of the normal modes of vibra-
tion are given by Eqs. (8-216), (8-222), and (8-223):
Wj mn
\Li x ,Li v Liz J
The three integers I, m, n cannot all be zero, for this gives w = and does
not correspond to a vibration of the fluid. If we combine these results
with Eqs. (8-217), (8-213), 8-211), and (8-207), we have for the normal
mode of vibration characterized by the numbers I, m, n:
i i a ^in- .a frra miry nirz , 00 _,
p = (A cos wi mn t + B sin Ul mn t) cos -f— cos -= — cos -f— > (8-225)
Lt x Liy '-'z
where we have suppressed the superfluous constant C x C y C z . The corre-
sponding velocities are
Vk , . . . t> ,s ■ tox miry nirz
V x = j (A Sin bilmnt — B COS U>l mn t) sin -j— COS -j — COS -j— >
J^xPo^lmn ^x Liy i-'z
m-K , . . *n ,\ It™ ■ fniry nirz
V y = j (A Sin Ulmnt — B cos o>i mn t) cos -j— sin —f — cos -=— >
LiyPcfUlmn Li x Lly Li z
nir , . . „ jX Itx rrvwy . nirz
Vz = J (A Sin Uimnt — B COS iOlmnt) COS -j- COS -= Sin -j— ■
L/zPo^lmn J-i x L/y Li z
(8-226)
These four equations give a complete description of the motion of the fluid
for a normal mode of vibration. The walls x = 0, x = L x , and the (I — 1)
equally spaced parallel planes between them are nodes for v x and anti-
nodes for p', v y , and v z . A similar remark applies to nodal planes parallel
to the other walls.
It will be observed that the normal frequencies are not, in general, har-
monically related to one another, as they were in the case of the vibrating
string. If, however, one of the dimensions, say L x , is much larger than the
other two, so that the box becomes a long square pipe, then the lowest
frequencies will correspond to the case where m = n = and I is a small
integer, and these frequencies are harmonically related. Thus, in a pipe,
the first few normal frequencies above the lowest will be multiples of the
lowest frequency. This explains why it is possible to get musical tones
from an organ pipe, as well as from a vibrating string. Our treatment here
applies only to a closed organ pipe, and a square one at that. The treat-
ment of a closed circular pipe is not much more difficult than the above
treatment and the general nature of the results is similar. The open ended
8-12] SOUND WAVES IN PIPES 341
pipe is, however, much more difficult to treat exactly. The difficulty lies
in the determination of the boundary condition at the open end; indeed,
not the least of the difficulties is in deciding just where the boundary is.
As a rough approximation, one may assume that the boundary is a plane
surface across the end of the pipe, and that this surface is a pressure node.
The results are then similar to those for the closed pipe, except that if one
end of a long pipe is closed and one open, the first few frequencies above
the lowest are all odd multiples of the lowest.
The general solution of the equations for sound vibrations in a rectangu-
lar cavity can be built up, as in the case of the vibrating string, by adding
normal mode solutions of the form (8-225) for all normal modes of vibra-
tion. The constants A and B for each mode of vibration can again be
chosen to fit the initial conditions, which in this case will be a specification
of p' and dp' /dt (or p' and v) at all points in the cavity at some initial
instant. We shall not carry out this development here. [In the above
discussion, we have omitted the case I = m = n = 0, which corresponds
to a constant pressure increment p' . Likewise, we omitted steady velocity
solutions v(z, y, z) which do not oscillate in time. These solutions would
have to be included in order to be able to fit all initial conditions.]
For cavities of other simple shapes, for example spheres and cylinders,
the method of separation of variables used in the above example works,
but in these cases instead of the variables x, y, z, coordinates appropriate
to the shape of the boundary surface must be used, for example spherical
or cylindrical coordinates. In most cases, except for a few simple shapes,
the method of separation of variables cannot be made to work. Approxi-
mate methods can be used when the shape is very close to one of the simple
shapes whose solution is known. Otherwise the only general methods of
solution are numerical methods which usually involve a prohibitive amount
of labor. It can be shown, however, that the general features of our re-
sults for rectangular cavities hold for all shapes; that is, there are normal
modes of vibration with characteristic frequencies, and the most general
motion is a superposition of these.
8-12 Sound waves in pipes. A problem of considerable interest is the
problem of the propagation of sound waves in pipes. We shall consider
a pipe whose axis is in the z-direction, and whose cross section is rectangu-
lar, of dimensions L x L y . This problem is the same as that of the preceding
section except that there are no walls perpendicular to the z-axis.
We shall apply the same method of solution, the only difference being
that the boundary conditions now apply only at the four walls x = 0,
x — L x , y = 0, y — L y . Consequently, we are restricted in our choice of
the functions X(x) and Y(y), just as in the preceding section, by Eqs.
(8-217), (8-222), and (8-223). There are no restrictions on our choice of
342 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
solution of the Z-equation (8-215). Since we are interested in solutions
representing the propagation of waves down the pipe, we choose the ex-
ponential form of solution for Z:
Z = e ik °*, (8-227)
and we choose the complex exponential solution (8-212) for @. Our solu-
tion for p', then, for a given choice of the integers I, m, is
JU X Ly
= A cos -j— cos -=r-^ cos (k z z — at). (8-228)
L x Ly
This represents a harmonic wave, traveling in the z-direction down the
pipe, whose amplitude varies over the cross section of the pipe according
to the first two cosine factors. Each choice of integers I, m corresponds
to what is called a mode of propagation for the pipe. (The choice I = 0,
m = is an allowed choice here.) For a given I, m and a given frequency w,
the wave number k z is determined by Eqs. (8-216), (8-222), and (8-223):
The plus sign corresponds to a wave traveling in the +z-direction, and
conversely. For I = m = 0, this is the same as the relation (8-201) for a
wave traveling with velocity c in the z-direction in a fluid filling three-
dimensional space. Otherwise, the wave travels with the velocity
Clm —
|fc.
-[-(^-(srr —>
which is greater than c and depends on w. There is evidently a minimum
frequency
Wim
[(#" + (")?'
below which no propagation is possible in the I, m mode; for k z would be
imaginary, and the exponent in Eq. (8-227) would be real, so that instead
of a wave propagation we would have an exponential decline in amplitude
of the wave in the z-direction. Note the similarity of these results to those
obtained in Section 8-4 for the discrete string, where, however, there was
an upper rather than a lower limit to the frequency. Since ci m depends on
<a, we again have the phenomenon of dispersion. A wave of arbitrary
8-13] THE MACH NUMBER 343
shape, which can be resolved into sinusoidally oscillating components of
various frequencies a, will be distorted as it travels along the pipe because
each component will have a different velocity. We leave as an exercise the
problem of calculating the fluid velocity v, and the power flow, associated
with the wave (8-228).
Similar results are obtained for pipes of other than rectangular cross
section. Analogous methods and results apply to the problem of the
propagation of electromagnetic waves down a wave guide. This is one
reason for our interest in the present problem.
8-13 The Mach number. Suppose we wish to consider two problems
in fluid flow having geometrically similar boundaries, but in which the
dimensions of the boundaries, or the fluid velocity, density, or compressi-
bility are different. For example, we may wish to investigate the flow of a
fluid in two pipes having the same shape but different sizes, or we may be
concerned with the flow of a fluid at different velocities through pipes of
the same shape, or with the flow of fluids of different densities. We might
be concerned with the relation between the behavior of an airplane and the
behavior of a scale model, or with the behavior of an airplane at different
altitudes, where the density of the air is different. Two such problems in-
volving boundaries of the same shape we shall call similar problems. Under
what conditions will two similar problems have similar solutions?
In order to make this question more precise, let us assume that for each
problem a characteristic distance s is defined which determines the geo-
metrical scale of the problem. In the case of similar pipes, s might be a
diameter of the pipe. In the case of an airplane, s might be the wing span.
We then define dimensionless coordinates x', y', z' by the equations
x' = x/s , y' = y/s 0) z' = z/s . (8-232)
The boundaries for two similar problems will have identical descriptions in
terms of the dimensionless coordinates x', y', z'\ only the characteristic
distance s will be different. In a similar way, let us choose a characteristic
speed v associated with the problem. The speed v Q might be the average
speed of flow of fluid in a pipe, or the speed of the airplane relative to the
stationary air at a distance from it, or v might be the maximum speed of
any part of the fluid relative to the pipe or the airplane. In any case, we
suppose that v is so chosen that the maximum speed of any part of the
fluid is not very much larger than v . We now define a dimensionless veloc-
ity v', and a dimensionless time coordinate t' :
v' = v/v , (8-233)
t' = v t/s . (8-234)
344 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
We now say that two similar problems have similar solutions if the solu-
tions are identical when expressed in terms of the dimensionless velocity v'
as a function of x', y', z', and t'. The fluid flow pattern will then be the
same in both problems, differing only in the distance and time scales de-
termined by s and v . We need also to assume a characteristic density
p and pressure p . In the case of the airplane, these would be the density
and pressure of the undisturbed atmosphere; in the case of the pipe, they
might be the average density and pressure, or the density and pressure at
one end of the pipe. We shall define a dimensionless pressure increment
p" as follows:
p" = P ~ Po • (8-235)
PoVo
We shall now assume that the changes in density of the fluid are small
enough so that we can write
9 = Po + % (P ~ Po) ' (8_236)
where higher order terms in the Taylor series for p have been neglected.
By making use of the definition (8-235) for p", and of the bulk modulus B
as given by Eq. (5-183), this can be written
p = p (l + M 2 p"), (8-237)
where
M =*°(p!r /2 =7- < 8 - 238)
Here M is the ratio of the characteristic velocity v to the velocity of sound
c and is called the Mach number for the problem. In a similar way, we can
expand 1/p, assuming that \p — p \ « p :
- = — (1 - M 2 p"). (8-239)
P Po
With the help of Eqs. (8-237) and (8-239), we can rewrite the equation
of continuity and the equation of motion in terms of the dimensionless
variables introduced by Eqs. (8-232) to (8-235). The equation of con-
tinuity (8-127), when we divide through by the constant po*>oAo and
collect separately the terms involving M , becomes
V'-v' + M 2 [^ + V- (p'V)J = 0, (8-240)
where
V ' = i l + >w + k is- < 8 - 241 >
8-14] viscosity 345
The equation of motion (8-139), when we divide through by t^/s , be-
comes, in the same way,
%■ + v'- VV + (1 - M 2 p") V'p" = ^ - • (8-242)
dt »o P
Equations (8-240) and (8-242) represent four differential equations to be
solved for the four quantities p', v', subject to given initial and boundary
conditions. If the body forces are zero, or if the body forces per unit mass
f/p are made proportional to vl/s , then the equations for two similar prob-
lems become identical if the Mach number M is the same for both. Hence,
similar problems will have similar solutions if they have the same Mach
number. Results of experiments on scale models in wind tunnels can be
extrapolated to full-sized airplanes flying at speeds with corresponding
Mach numbers. If the Mach number is much less than one, the terms in
M 2 in Eqs. (8-240) and (8-242) can be neglected, and these equations then
reduce to the equations for an incompressible fluid, as is obvious either
from Eq. (8-240) or (8-237). Therefore at fluid velocities much less than
the speed of sound, even air may be treated as an incompressible fluid.
On the other hand, at Mach numbers near or greater than one, the com-
pressibility becomes important, even in problems of liquid flow. Note that
the Mach number involves only the characteristic velocity v , and the
velocity of sound, which in turn depends on the characteristic density p
and the compressibility B. Changes in the distance scale factor s have no
effect on the nature of the solution, nor do changes in the characteristic
pressure p except insofar as they affect p and B.
It must be emphasized that these results are applicable only to ideal
fluids, i.e., when viscosity is unimportant, and to problems where the den-
sity of the fluid does not differ greatly at any point from the characteristic
density p . The latter condition holds fairly well for liquids, except when
there is cavitation (formation of vapor bubbles), and for gases except at
very large Mach numbers.
8-14 Viscosity. In many practical applications of the theory of fluid
flow, it is not permissible to neglect viscous friction, as has been done in
the preceding sections. When adjacent layers of fluid are moving past one
another, this motion is resisted by a shearing force which tends to reduce
their relative velocity. Let us assume that in a given region the velocity
of the fluid is in the z-direction, and that the fluid is flowing in layers
parallel to the zz-plane, so that v x is a function of y only (Fig. 8-10).
Let the positive y-axis be directed toward the right. Then if dv x /dy is
positive, the viscous friction will result in a positive shearing force F x
acting from right to left across an area A parallel to the xz-plane. The
346
THE MECHANICS OF CONTINUOUS MEDIA
[CHAP. 8
Fig. 8-10. Velocity distribution in the definition of viscosity.
coefficient of viscosity r\ is defined as the ratio of the shearing stress to the
velocity gradient :
F x /A
dv x /dy
(8-243)
When the velocity distribution is not of this simple type, the stresses due
to viscosity are more complicated. (See Section 10-6.)
We shall apply this definition to the important special case of steady
flow of a fluid through a pipe of circular cross section, with radius a. We
shall assume laminar flow; that is, we shall assume that the fluid flows in
layers, as contemplated in the definition above. In this case, the layers
are cylinders. The velocity is everywhere parallel to the axis of the pipe,
which we take to be the z-axis, and the velocity v z is a function only of r,
the distance from the axis of the pipe. (See Fig. 8-11.) If we consider a
cylinder of radius r and of length I, its area will be A = 2-irrl, and accord'
ing to the definition (8-243), the force exerted across this cylinder by the
fluid outside on the fluid inside the cylinder is
F.=
dv z
*CM>¥"
(8-244)
Since the fluid within this cylinder is not accelerated, if there is no body
force the viscous force must be balanced by a difference in pressure
8-14]
VISCOSITY
347
Fig. 8-1 1 . Laminar flow in a pipe.
between the two ends of the cylinder:
Ap{irr 2 ) + F, = 0,
(8-245)
where Ap is the difference in pressure between the two ends of the cylinder
a distance I apart, and we assume that the pressure is uniform over the
cross section of the pipe. Equations (8-244) and (8-245) can be combined
to give a differential equation for v z :
dv z
dr
r Ap
2 v l '
(8-246)
We integrate outward from the cylinder axis :
L dv >=-wL rdr >
Vz = V
r 2 Ap
(8-247)
where v is the velocity at the axis of the pipe. We shall assume that the
fluid velocity is zero at the walls of the pipe :
r i ° 2 A P n
[Vz] r =a = Vq -^- = 0,
(8-248)
although this assumption is open to question.
348 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8
Then
(8-249)
tT/fr
and
Vo =
a 2 Ap
4ul
z
Ap
(a 2 -
r 2 )
(8-250)
The total fluid current through the pipe is
I = Jjpv z dS = 2wp J" v z r dr. (8-251;
We substitute from Eq. (8-250) and carry out the integration :
This formula is called Poiseuille's law. It affords a convenient and simple
way of measuring i\.
Although we will not develop now the general equations of motion for
viscous flow, we can arrive at a result analogous to that in Section 8-13,
taking viscosity into account, without actually setting up the equations
for viscous flow. Suppose that we are concerned, as in Section 8-13, with
two similar problems in fluid flow, and let s , v , p , p be a characteristic
distance, velocity, pressure, and density, which again define the scale in
any problem. However, let us suppose that in this case viscosity is to be
taken into account, so that the equation of motion (8-139) is augmented
by a term corresponding to the force of viscous friction. We do not at
present know the precise form of this term, but at any rate it will consist
of rj multiplied by various derivatives of various velocity components,
and divided by p [since Eq. (8-139) has already been divided through
by p]. When we introduce the velocity v', and the dimensionless coordi-
nates x', y', z', t', as in Section 8-13, and divide the equation of motion
by v\/s , we will obtain just Eq. (8-242), augmented by a term involving
the coefficient of viscosity. Since all the terms in Eq. (8-242) are dimen-
sionless, the viscosity term will be also, and will consist of derivatives of
components of v' with respect to x', y', z', multiplied by numerical factors
and by a dimensionless coefficient consisting of rj times some combination
of vo and s , and divided by p = p (l + M 2 p") [Eq. (8-237)]. Now the
dimensions of r\, as determined by Eq. (8-243), are
^ = length a xtime' ( 8 " 253 >
and the only combination of p , v , and s having these dimensions
8-14] viscosity 349
is PoVqS - Therefore the viscosity term will be multiplied by the coeffi-
cient
1
R(l + M 2 p")
where R is the Reynolds number, defined by
(8-254)
^ = M o£o_ (8255)
v
We can now conclude that when viscosity is important, two similar prob-
lems will have the same equation of motion in dimensionless variables, and
hence similar solutions, only if the Reynolds number R, as well as the
Mach number M, is the same for both. If the Mach number is very small,
then compressibility is unimportant. If the Reynolds number is very large,
then viscosity may be neglected. It turns out that there is a critical value
of Reynolds number for any given problem, such that the nature of the
flow is very different for R larger than this critical value than for smaller
values of R. For small Reynolds numbers, the flow is laminar, as the
viscosity tends to damp out any vortices which might form. For large
Reynolds numbers, the flow tends to be turbulent. This will be the case
when the viscosity is small, or the density, velocity, or linear dimensions
are large. Note that the Reynolds number depends on s , whereas the
Mach number does not, so that the distance scale of a problem is im-
portant when the effects of viscosity are considered. Viscous effects are
more important on a small scale than on a large scale.
It may be noted that the expression (8-255) for the Reynolds number,
together with the fact that Eq. (8-139) is divided by vo/s to obtain the
dimensionless equation of motion, implies that the viscosity term to be
added to Eq. (8-139) has the dimensions of (nD )/(p sl). This, in turn,
implies that the viscous force density must be equal to 17 times a sum of
second derivatives of velocity components with respect to x, y, and z.
This is perhaps also evident from Eq. (8-243), since in calculating the
total force on a fluid element, the differences in stresses on opposite faces
of the element will be involved, and hence a second differentiation of
velocities relative to x, y, and z will appear in the expression for the force.
An expression for the viscous force density will be developed in Chapter 10.
350 the mechanics of continuous media [chap. 8
Problems
1. A stretched string of length I is terminated at the end z = I by a ring of
negligible mass which slides without friction on a vertical rod. (a) Show that the
boundary condition at this end of the string is
[-1 =
0.
(b) If the end x = is tied, find the normal modes of vibration.
2. Find the boundary condition and the normal modes of vibration in Prob-
lem 1 if the ring at one end has a finite mass m. What is the significance of the
limiting cases m = and m = «> ?
3. The midpoint of a stretched string of length I is pulled a distance u — Z/10
from its equilibrium position, so that the string forms two legs of an isosceles
triangle. The string is then released. Find an expression for its motion by the
Fourier series method.
4. A piano string of length I, tension t, and density a, tied at both ends, and
initially at rest, is struck a blow at a distance o from one end by a hammer of
mass m and velocity Do. Assume that the hammer rebounds elastically with
velocity — vo, and that its momentum loss is transferred to a short length Al
of string centered around x = a. Find the motion of the string by the Fourier
series method, assuming that Al is negligibly small. If the finite length of Al
were taken into account, what sort of effect would this have on your result? If
it is desired that no seventh harmonic of the fundamental frequency be present
(it is said to be particularly unpleasant), at what points a may the string be
struck?
5. A string of length I is tied at x = I. The end at x = is forced to move
sinusoidally so that
m(0, t) = A sin <iit.
(a) Find the steady-state motion of the string; that is, find a solution in which
all points on the string vibrate with the same angular frequency u. (b) How
would you find the actual motion if the string were initially at rest?
6. A force of linear density
,, .-. , . nwx
f(x, t) = /o sin — j— cos cot,
where n is an integer, is applied along a stretched string of length I. (a) Find
the steady-state motion of the string. [Hint: Assume a similar time and space
dependence for u(x, t), and substitute in the equation of motion.] (b) Indicate
how one might solve the more general problem of a harmonic applied force
f&y t) = fo(x) cos ut,
where fo(x) is any function.
7. Assume that the friction of the air around a vibrating string can be repre-
sented as a force per unit length proportional to the velocity of the string. Set
PROBLEMS 351
up the equation of motion for the string, and find the normal modes of vibration
if the string is tied at both ends.
8. Find the motion of a horizontal stretched string of tension t, density cr,
and length I, tied at both ends, taking into account the weight of the string. The
string is initially held straight and horizontal, and dropped. [Hint: Find the
steady-state "motion" and add a suitable transient.]
9. A long string is terminated at its right end by a massless ring which slides
on a vertical rod and is impeded by a frictional force proportional to its velocity.
Set up a suitable boundary condition and discuss the reflection of a wave at the
end. How does the reflected wave behave in the limiting cases of very large and
very small friction? For what value of the friction constant is there no reflected
wave?
10. Discuss the reflection of a wave traveling down a long string terminated
by a massless ring, as in Problem 1.
11. Find a solution to Problem 3 by superposing waves f(x — ct) and
g(x + ct) in such a way as to satisfy the initial and boundary conditions. Sketch
the appearance of the string at times t = 0, %1/c, \l/c, and l/c.
12. (a) A long stretched string of tension t and density <n is tied at x =
to a string of density 02- If the mass of the knot is negligible, show that u
and du/dx must be the same on both sides of the knot.
(b) A wave A cos {k\x — ojt) traveling toward the right on the first string
is incident on the junction. Show that in order to satisfy the boundary conditions
at the knot, there must be a reflected wave traveling to the left in the first string
and a transmitted wave traveling to the right in the second string, both of the
same frequency as the incident wave. Find the amplitudes and phases of the
incident and reflected waves.
(c) Check your result in part (b) by calculating the power in the transmitted
and reflected waves, and showing that the total is equal to the power in the
incident wave.
13. Derive directly from Eq. (8-139) an equation expressing the conservation
of angular momentum in a form analogous to Eq. (8-141).
14. Derive an equation expressing the law of conservation of angular mo-
mentum for a fluid in a form analogous to Eq. (8-140). From this, derive equa-
tions analogous to Eqs. (8-141), (8-142), and (8-144). Explain the physical
meaning of each term in each equation. Show that the internal torques due to
pressure can be eliminated from the integrated forms, and derive an equation
analogous to Eq. (8-148).
15. Derive and interpret the following equation:
it jjj {ipt)2 ~ p9 ~ pu) dV + IS n ' v( * p " 2 ~~ p9 ~ pu) dS
dt
v
j P ^ dV >
S V
where 7 is a fixed volume bounded by a surface S with normal n.
352 THE MECHANICS OP CONTINUOUS MEDIA [CHAP. 8
16. (a) A mass of initially stationary air at 45° N latitude flows inward toward
a low-pressure spot at its center. Show that the coriolis torque about the low-
pressure center depends only on the radial component of velocity. Hence, show
that if frictional torques are neglected, the angular momentum per unit mass at
radius r from the center depends only on r and on the initial radius ro at which
the air is stationary, but does not depend on the details of the motion.
(b) Calculate the azimuthal component of velocity around the low as a func-
tion of initial and final radius. If this were a reasonable model of a tornado,
what would be the initial radius ro if the air at 264 ft from the center has a
velocity of 300 mi/hr?
17. Evaluate the potential energy u per unit mass as a function of p for a
perfect gas of molecular weight M at temperature T. For the steady isothermal
flow of this gas through a pipe of varying cross section and varying height above
the earth, find expressions for the pressure, density, and velocity of the gas as
functions of the cross section S of the pipe, the height h, and the pressure po
and velocity vq at a point in the pipe at height h = where the cross section
is So- Assume p, v, and p uniform over the cross section.
18. Work Problem 17 for an incompressible fluid of density po-
19. The function <t> = a/r, where a is a constant and r is the distance from a
fixed point, satisfies Laplace's equation (8-177), except at r = 0, because it has
the same form as the gravitational potential of a point mass. If this is a velocity
potential, what is the nature of the fluid flow to which it leads?
20. (a) Verify by direct computation that the spherical wave (8-204) satisfies
the wave equation (8-186). (b) Write an analogous expression for a cylindrical
wave of arbitrary time dependence, traveling out from the z-axis, independent
of z and with cylindrical symmetry. Make the amplitude depend on the dis-
tance from the axis in such a way as to satisfy the requirement of conservation
of energy. Show that such a wave cannot satisfy the wave equation. (It is a
• general property of cylindrical waves that they do not preserve their shape.)
*21. Show that the normal mode of vibration given by Eqs. (8-225) and (8-226)
can be represented as a superposition of harmonically oscillating plane waves
traveling in appropriately chosen directions with appropriate phase relation-
ships. Show that in the normal vibrations of a fluid in a box, the velocity oscil-
lates 90° out of phase with the pressure at any point. How can this be reconciled
with the fact that in a plane wave the velocity and pressure are in phase?
22. Find the normal modes of vibration of a square organ pipe with one end
open and the other closed, on the assumption that the open end is a pressure node.
23. (a) Calculate the fluid velocity v for the wave given by Eq. (8-228). (b)
Calculate the mean rate of power flow through the pipe.
*24. Show that the expression (8-228) for a sound wave in a pipe can be repre-
sented as a superposition of plane waves traveling with speed c in appropriate
directions, and being reflected at the walls. Explain, in terms of this representa-
tion, why there is a minimum frequency for any given mode below which a wave
cannot propagate through the pipe in this mode.
25. If the sound wave given by Eq. (8-228) is incident on a closed end of the
pipe at z = 0, find the reflected wave.
PROBLEMS 353
26. Develop the theory of the propagation of sound waves in a circular pipe,
using cylindrical coordinates and applying the method of separation of variables.
Carry the solution as far as you can. You are not required to solve the equation
for the radial part of the wave, but you should indicate the sort of solutions you
would expect to find.
27. A fluid of viscosity i\ flows steadily between two infinite parallel plane
walls a distance I apart. The velocity of the fluid is everywhere in the same direc-
tion, and depends only on the distance from the walls. The total fluid current
between the walls in any unit length measured along the walls perpendicular to
the direction of flow is I. Find the velocity distribution and the pressure gradient
parallel to the walls, assuming that the pressure varies only in the direction of
flow.
28. Prove that the only combination of po, fo, so having the dimensions of
viscosity is po^oso-
CHAPTER 9
LAGRANGE'S EQUATIONS
9-1 Generalized coordinates. Direct application of Newton's laws to
a mechanical system results in a set of equations of motion in terms of the
cartesian coordinates of each of the particles of which the system is com-
posed. In many cases, these are not the most convenient coordinates in
terms of which to solve the problem or to describe the motion of the system.
For example, in the problem of the motion of a single particle acted on by
a central force, which we treated in Section 3-13, we found it convenient to
introduce polar coordinates in the plane of motion of the particle. The
reason was that the force in this case can be expressed more simply in terms
of polar coordinates. Again in the two-body problem, treated in Section
4-7, we found it convenient to replace the coordinates ri, r 2 of the two
particles by the coordinate vector R of the center of mass, and the relative
coordinate vector r which locates particle 1 with respect to particle 2. We
had two reasons for this choice of coordinates. First, the mutual forces
which the particles exert on each other ordinarily depend on the relative
coordinate. Second, in many cases we are interested in a description of
the motion of one particle relative to the other, as in the case of planetary
motion. In problems involving many particles, it is usually convenient
to choose a set of coordinates which includes the coordinates of the center
of mass, since the motion of the center of mass is determined by a relatively
simple equation (4-18). In Chapter 7, we found the equations of motion
of a particle in terms of moving coordinate systems, which are sometimes
more convenient to use than the fixed coordinate systems contemplated
in Newton's original equations of motion.
We shall include coordinate systems of the sort described above, to-
gether with cartesian coordinate systems, under the name generalized co-
ordinates. A set of generalized coordinates is any set of coordinates by
means of which the positions of the particles in a system may be specified.
In a problem requiring generalized coordinates, we may set up Newton's
equations of motion in terms of cartesian coordinates, and then change to
the generalized coordinates, as in the problems studied in previous chap-
ters. It would be very desirable and convenient, however, to have a gen-
eral method for setting up equations of motion directly in terms of any
convenient set of generalized coordinates. Furthermore it is desirable to
have uniform methods of writing down, and perhaps of solving, the
equations of motion in terms of any coordinate system. Such a method
was invented by Lagrange and is the subject of this chapter.
354
9-1] GENERALIZED COORDINATES 355
In each of the cases mentioned in the first paragraph, the number of
coordinates in the new system of coordinates introduced to simplify the
problem was the same as the number of cartesian coordinates of all the
particles involved. We may, for example, replace the two cartesian co-
ordinates x, y of a particle moving in a plane by the two polar coordinates
r, 0, or the three space coordinates x, y, z by three spherical or cylindrical
coordinates. Or we may replace the six coordinates x lt y x , z lt x 2 , y 2 , z 2
of a pair of particles by the three coordinates X, Y, Z of the center of
mass plus the three coordinates x, y, z of one particle relative to the other.
Or we may replace the three coordinates of a particle relative to a fixed
system of axes by three coordinates relative to moving axes. (A vector
counts as three coordinates.)
In our treatment of the rotation of a rigid body about an axis (Section
5-2), we described the position of the body in terms of the single angular
coordinate 6. Here we have a case where we can replace a great many
cartesian coordinates, three for each particle in the body, by a single
coordinate 0. This is possible because the body is rigid and is allowed to
rotate only about a fixed axis. As a result of these two facts, the position
of the body is completely determined when we specify the angular position
of some reference line in the body. The position of a free rigid body can
be specified by six coordinates, three to locate its center of mass, and three
to determine its orientation in space. This is a vast simplification com-
pared with the 3N cartesian coordinates required to locate its N particles.
A rigid body is an example of a system of particles subject to constraints,
that is, conditions which restrict the possible sets of values of the coordi-
nates. In the case of a rigid body, the constraint is that the distance
between any two particles must remain fixed. If the body can rotate
only about a fixed axis, then in addition the distance of each particle from
the axis is fixed. This is the reason why specifying the value of the single
coordinate is sufficient to determine the position of each particle in the
body. We shall postpone the discussion of systems like this which involve
constraints until Section 9-4. In this section, and the next, we shall set
up the theory of generalized coordinates, assuming that there are as many
generalized coordinates as cartesian coordinates. We shall then find, in
Section 9-4, that this theory applies also to the motion of constrained
systems.
When we want to speak about a physical system described by a set of
generalized coordinates, without specifying for the moment just what the
coordinates are, it is customary to designate each coordinate by the letter q
with a numerical subscript. A set of n generalized coordinates would be
written as q lt q 2 , . . . , q n - Thus a particle moving in a plane may be
described by two coordinates q u q 2 , which may in special cases be the
cartesian coordinates x, y, or the polar coordinates r, 0, or any other suit-
356
lagrange's equations
[chap. 9
able pair of coordinates. A particle moving in space is located by three
coordinates, which may be cartesian coordinates x, y, z, or spherical
coordinates r, 0, <p, or cylindrical coordinates p, z, <p, or, in general
5l, «2, ?3-
The configuration of a system of TV particles may be specified by the
37V cartesian coordinates xi, y\, Z\, x 2 , y 2 , z 2 , ■ ■ ■ , xx, y^, z^ of its par-
ticles, or by any set of 37V generalized coordinates q u q 2 , . . . , q3N- Since
for each configuration of the system, the generalized coordinates must
have some definite set of values, the coordinates g 1( . . . , q3w will be func-
tions of the cartesian coordinates, and possibly also of the time in the
case of moving coordinate systems:
Qi = 2i(zi, 2/i, zi, x 2 , 2/2, • • ■ , Vn, z n ; t),
?2 = 92(^1, 2/i, , ZN)t),
13N = q3N{Xl, 2/1, , z N ;t).
(9-1)
Since the coordinates q\, . . . , qz n specify the configuration of the system,
it must be possible also to express the cartesian coordinates in terms of
the generalized coordinates:
Xi = xi(qi, q 2 , ■ ■ ■ ,Q3n; t),
2/1 = 2/1(91, , Q3n; t),
zn = zjv(<7i, , Q3N', t).
(9-2)
If Eqs. (9-1) are given, they may be solved for x\, 2/1, . . . , zjv to obtain
Eqs. (9-2), and vice versa.
The mathematical condition that this solution be (theoretically) possible is
that the Jacobian determinant of Eqs. (9-1) be different from zero at all points,
or nearly all points:
dgi bq 2 dq3N
dxi dxi dxi
d(qi, . . . , g3iv)
d(xi, 3/i,
, zjv)
dgi
dq 2
dyi
dq3N
dyi
dgi
dztf
dq 2
dq3N
dzif
j* 0.
(9-3)
If this inequality does not hold, then Eqs. (9-1) do not define a legitimate set
of generalized coordinates. In practically all cases of physical interest, it will
9-1] GENERALIZED COORDINATES 357
be evident from the geometrical definitions of the generalized coordinates
whether or not they are a legitimate set of coordinates. Thus we shall not
have any occasion to apply the above test to our coordinate systems. [For a
derivation of the condition (9-3), see W. F. Osgood, Advanced Calculus, New
York: Macmillan, 1937, p. 129.]
As an example, we have the equations (3-72) and (3-73) connecting
the polar coordinates r, 6 of a single particle in a plane with its cartesian
coordinates x, y. As an example of a moving coordinate system, we
consider polar coordinates in which the reference axis from which 6 is
measured rotates counterclockwise with constant angular velocity w
(Fig. 9-1):
r=(z 2 + t/ 2 ) 1/2 ,
6 = tan -1 ^ — at, (9-4)
and conversely,
x — r cos (0 + ut),
y = rsin (0 + ut). (9-5)
As an example of generalized coordinates for a system of particles, we
have the center of mass coordinates X, Y, Z and relative coordinates
x, y, z of two particles of masses m l and m 2 , as denned by Eqs. (4-90)
and (4-91), where X, Y, Z are the components of R, and x, y, z are the
components of r. Because the transformation equations (4-90) and
(4-91) do not contain the time explicitly, we regard this as a fixed co-
ordinate system, even though x, y, z are the coordinates of mi referred
to a moving origin located on m 2 . The rule which defines the coordinates
X, Y, Z, x, y, z is the same at all times.
If a system of particles is described by a set of generalized coordinates
flii • • • > Q3N, we shall call the time derivative q k , of any coordinate <?&,
the generalized velocity associated with this coordinate. The generalized
velocity associated with a cartesian coordinate a;,- is just the correspond-
Fig. 9-1. A rotating polar coordinate system.
358 lagrange's equations [chap. 9
ing component Xi of the velocity of the particle located by that coordinate.
The generalized velocity associated with an angular coordinate is the
corresponding angular velocity 0. The velocity associated with the co-
ordinate X in the preceding example is X, the x-component of velocity
of the center of mass. The generalized velocities can be computed in terms
of cartesian coordinates and velocities, and conversely, by differentiating
Eqs. (9-1) or (9-2) with respect to t according to the rules for differentiat-
ing implicit functions. For example, the cartesian velocity components
can be expressed in terms of the generalized coordinates and velocities
by differentiating Eqs. (9-2):
¥k dxi . . dxi
i 0-6)
dZN . i d*N
As an example, we have, from Eqs. (9-5):
x = f cos (0 + at) — r0 sin (0 + at) — m sin (0 + ait),
(9—7)
y = r sin (0 + at) + rO cos (0 + at) + ru cos (0 + at).
The kinetic energy of a system of N particles, in terms of cartesian
coordinates, is
T = j^ *»»<(*? + y' + *?)■ (»-8)
By substituting from Eqs. (9-6), we obtain the kinetic energy in terms
of generalized coordinates. If we rearrange the order of summation, the
result is
fc=l 1=1 k=l
where
9-1] GENEEALIZED COORDINATES 359
The coefficients Am, B^, and T§ are functions of the coordinates
<h> • • • > Izn, and also of t for a moving coordinate system. If A k i is
zero except when k = I, the coordinates are said to be orthogonal. The coef-
ficients B k and T Q are zero when t does not occur explicitly in Eqs. (9-1),
i.e., when the generalized coordinate system does not change with time.
We see that the kinetic energy, in general, contains three sets of terms:
T = T 2 + T x + T , (9-13)
where T 2 contains terms quadratic in the generalized velocities, Ti con-
tains linear terms, and T is independent of the velocities. The terms
T x and T appear only in moving coordinate systems; for fixed coordinate
systems, the kinetic energy is quadratic in the generalized velocities.
As an example, in plane polar coordinates [Eqs. (3-72)], the kinetic
energy is
T = %m(x 2 + y 2 )
= \{mf 2 + mr 2 6 2 ), (9-14)
as may be obtained by direct substitution from Eqs. (3-72), or as a
special case of Eq. (9-9), where
dx doc
-=cos0, -= -rsinfl,
dy ■ » dy
f r = sme, f e = rco S e.
(9-15)
If we take the moving coordinate system defined by Eqs. (9-5), we find,
by substituting from Eqs. (9-7), or by using Eq. (9-9),
T = £m(x 2 + V 2 )
= i(mf 2 + mr 2 6 2 ) + mr 2 o)6 + %mr 2 w 2 . (9-16)
In this case, a term linear in 6 and a term independent of f and 6 appear.
The kinetic energy for the two-particle system can also easily be written
down in terms of X, Y, Z, x, y, z, defined by Eqsi (4-90) and (4-91).
Instead of finding the kinetic energy first in cartesian coordinates and
then translating into generalized coordinates, as in the examples above,
it is often quicker to work out the kinetic energy directly in terms of
generalized coordinates from a knowledge of their geometrical meaning.
It may then be possible to start a problem from the beginning with a
suitable set of generalized coordinates without writing out explicitly the
transformation equations (9-1) and (9-2) at all. For example, we may
obtain Eq. (9-14) immediately from the geometrical meaning of the
360 lagrange's equations [chap. 9
coordinates r, (see Fig. 3-20) by noticing that the linear velocity associ-
ated with a change in r is f and that associated with a change in is r&.
Since the directions of the velocities associated with r and are perpen-
dicular, the square of the total velocity is
v 2 = r 2 + r 2 8 2 , (9-17)
from which Eq. (9-14) follows immediately.
Care must be taken in applying this method if the velocities associated
with changes of the various coordinates are not perpendicular. For exam-
ple, let us consider a pair of coordinate axes u, w making an angle a less
than 90° with each other, as in Fig. 9-2. Let u and w be the sides of a par-
allelogram formed by these axes and by lines parallel to the axes through
the mass m as shown. Let a and b be unit vectors in the directions of
increasing u and w. Using u and w as coordinates, the velocity of the
mass m is
v = wa + wb. (9-18)
The kinetic energy is
T = jtov-v = \mu 2 + \mw 2 + muw cos a. (9-19)
This is an example of a set of nonorthogonal coordinates in which a cross
product term in the velocities appears in the kinetic energy. The reason
for using the term orthogonal, which means perpendicular, is clear from
this example.
When systems of more than one particle are described in terms of
generalized coordinates, it is usually safest to write out the kinetic energy
first in cartesian coordinates and transform to generalized coordinates.
However, in some cases, it is possible to write the kinetic energy directly
in general coordinates. For exam-
ple, if a rigid body rotates about an
axis, we know that the kinetic
energy is J/w 2 , where w is the angu-
lar velocity about that axis and / is
the moment of inertia. Also, we can
use the theorem proved in Section
4-9 that the total kinetic energy of
a system of particles is the kinetic
energy associated with the center of
mass plus that associated with the
internal coordinates. [See Eq.
Fig. 9-2. A nonorthogonal coordi- (4-127).] As an example, the kinetic
nate system. energy of the two-particle system in
9-1] GENERALIZED COORDINATES 361
terms of the coordinates X, Y, Z, x, y, z, defined by Eqs. (4-90) and (4-91)
is
T = W(X 2 + f 2 + Z 2 ) + i M (x 2 + y 2 + z 2 ), (9-20)
where M and n are given by Eqs. (4-97) and (4-98). The result shows
that this is an orthogonal coordinate system. If the linear velocity of
each particle in a system can be written down directly in terms of the
generalized coordinates and velocities, then the kinetic energy can im-
mediately be written down.
We now note that the components of the linear momentum of particle i,
according to Eq. (9-8), are
dT dT . dT , n .,,
p ix = mxi = — , p iy = myi = — > Piz = mz t = — • (9-21)
aXi dyi dZi
In the case of a particle moving in a plane, the derivatives of T with
respect to f and 6, as given by Eq. (9-14), are
p r = m r= —, Ve = mr 2 e = -^ , (9-22)
where p r is the component of linear momentum in the direction of in-
creasing r, and p e is the angular momentum about the origin. Similar
results will be found for spherical and cylindrical coordinates in three
dimensions. In fact, it is not hard to show that for any coordinate q^
which measures the linear displacement of any particle or group of par-
ticles in a given direction, the linear momentum of that particle or group
in the given direction is dT/dqk) and that for any coordinate q k which
measures the angular displacement of a particle or group of particles
about an axis, their angular momentum about that axis is dT/dq k . This
suggests that we define the generalized momentum pk associated with the
coordinate q^ by*
a/77
p k = Wk - (9-23)
If qk is a distance, pk is the corresponding linear momentum. If q k is an
* The kinetic energy T is denned by Eq. (9-9) as a function of qi, . . . , J3at;
gi, . . . , q3N, and perhaps of t. The derivatives of this function T with respect
to these variables will be denoted by the symbols for partial differentiation.
Since qi, . . . , qw\ ?i, . . . , qsN are all functions of the time t for any given
motion of the system, T is also a function of t alone for any given motion. The
derivative of T with respect to time in this sense will be denoted by d/dt. The
same remarks apply to any other quantity which may be written as a function
of the coordinates and velocities and perhaps of t, and which is also a function
of t alone for any given motion.
362 lagrange's equations [chap. 9
angle, pk is the corresponding angular momentum. In other cases, pk
will have some other corresponding physical significance. According to
Eq. (9-9), the generalized momentum pk is
Pk = X) A "ih + B k- (9-24)
i=i
In the case of the coordinates X, Y, Z, x, y, z for the two-particle sys-
tem, this definition gives
p x = MX, p Y = MY, p z = MZ,
p x = fix, p v = ny, p z = nz, (9-25)
where px , Py, Pz are the components of the total linear momentum of
the two particles, and p x , p y , p z are the linear momentum components
in the equivalent one-dimensional problem in x, y, z to which the two-
body problem was reduced in Section 4-7. We shall see in the next sec-
tion that the analogy between the generalized momenta pk and the
cartesian components of linear momentum can be extended to the equa-
tions of motion in generalized coordinates.
If forces Fix, F Xy , F u , . . . , Fnz act on the particles, the work done
by these forces if the particles move from the positions x\, y x , z x , . . . , zm
to nearby points X\ + hx x , y x + 6y lt Z\ + hz\, . . . , z N + 8zn is
iv
SW = X) (*"<* Sxi + F iv 8y { + F iz to,). (9-26)
i=i
The small displacements Sx^ 8y{, 8zt may be expressed in terms of gen-
eralized coordinates:
^ dxi $
where 8qi, . . . , 5g 3 jv are the differences in the generalized coordinates
associated with the two sets of positions of the particles. We call this
a virtual displacement of the system because it is not necessary that it
represent any actual motion of the system. It may be any possible mo-
tion of the system. In the case of a moving coordinate system, we regard
the time as fixed; that is, we specify the changes in position in terms of
9-1] GENERALIZED COORDINATES 363
the coordinate system at a particular time t. If we substitute Eqs. (9-27)
in Eq. (9-26), we have, after rearranging terms:
32V
SW = 2 Qk dq k , (9-28)
fc=i
where
The coefficients Q k depend on the forces acting on the particles, on the
coordinates q\, . . . , q SN , and possibly also on the time t. In view of the
similarity in form between Eqs. (9-26) and (9-28), it is natural to call
the quantity Q k the generalized force associated with the coordinate q k .
We can define the generalized force Q k directly, without reference to the
cartesian coordinate system, as the coefficient which determines the
work done in a virtual displacement in which q k alone changes:
SW = Q k Sq k , (9-30)
where SW is the work done when the system moves in such a way that
q k increases by Sq k , all other coordinates remaining constant. Notice
that the work in Eq. (9-26), and therefore also in Eq. (9-30), is to be
computed from the values of the forces for the positions x x , . . . , z N , or
Qi> • • • , qzN) that is, we do not take account of any change in the forces
during the virtual displacement.
If the forces F lx , . . . , F Nz are derivable from a potential energy
V(x u ..., z N ) [Eqs. (4-32)], then
SW = -SV
If V is expressed in terms of generalized coordinates, then
SW = —SV
3N r)V
By comparing this with Eq. (9-28), we see that
Q*=-g, (9-33)
which shows that in this sense also the definition of Q k as a generalized
364 lagrange's equations [chap. 9
force is a natural one. Equation (9-33) may also be verified by direct
calculation of dV/dq k :
dV = y. /W dxi dVdyt dV dzA
dq k ~~ £-J \dxi dq k dyt dq k dz» dqj
= -Q k .
As an example, let us calculate the generalized forces associated with
the polar coordinates r, 0, for a particle acted on by a force
F = iF x + W v = nF r + lF e . (9-34)
If we use the definition (9-29), we have, using Eqs. (9-15) :
= F x cos + Fy sin
= F r , (9-35)
_ dx dy
Qe ~ * x dd + tv de
= —rF x sin 6 + rF v cos
= rF,.
We see that Q r is the component of force in the r-direction, and Q e is
the torque acting to increase 0. It is usually quicker to use the defini-
tion (9-30), which enables us to bypass the cartesian coordinates altogether.
If we consider a small displacement in which r changes to r + 5r, with 6
remaining constant, the work is
SW = F r 5r, (9-36)
from which the first of Eqs. (9-35) follows. If we consider a displace-
ment in which r is fixed and increases by 50, the work is
SW = F e r SO, (9-37)
from which the second of Eqs. (9-35) follows. In general, if q k is a co-
ordinate which measures the distance moved by some part of the mechani-
cal system in a certain direction, and if F k is the component in this direc-
tion of the total force acting on this part of the system, then the work done
9-2] lagbange's equations 365
when q k increases by dq k , all other coordinates remaining constant, is
SW = F k Sq k . (9-38)
Comparing this with Eq. (9-30), we have
Qk = Ft. (9-39)
In this case, the generalized force Q k is just the ordinary force F k . If q k
measures the angular rotation of a certain part of the system about a
certain axis, and if N k is the total torque about that axis exerted on this
part of the system, then the work done when q k increases by Sq k is
SW = N k 8q k . (9-40)
Comparing this with Eq. (9-30), we have
Qk = N k . (9-4:1)
The generalized force Q k associated with an angular coordinate q k is the
corresponding torque.
9-2 Lagrange's equations. The analogy which led to the definitions of
generalized momenta and generalized forces tempts us to suspect that
the generalized equations of motion will equate the time rate of change
of each momentum p k to the corresponding force Q k . To check this
suspicion, let us calculate the time rate of change of p k :
We will need to start with Newton's equations of motion in cartesian form:
rriiVi = Fiy, [i = 1, . . . , N] (9^3)
rtii'Zi = F iz .
Therefore we express T in cartesian coordinates [Eq. (9-8)]. We then
have
dT
dq k
— = >,m I la;,TT i + Vi -^ *■ + & t — * I , (9-44)
where x u y u . . . , z N are given as functions of q u . . . , q^N', ii, ■ • ■ , q>3N; t
by Eqs. (9-6). Since dXi/dq k and dXi/dt are functions only of q u . . . , q 3 ^;
t, we have, by differentiating Eqs. (9-6) :
366 lagrange's equations [chap. 9
dii dXi
dqk dq k '
W = d £'~ H=l,--.,N;k=l,...,M] (9-45)
oq k °qk
dZj _ dzj
dqk dq k
By substituting from Eqs. (9-45) in Eq. (9~44), and differentiating
again with respect to t, we obtain
dt f{ \ dq k a dq k dq k /
. v-> ( . d dx{ . . d dyi . . d dzi\ /n .„.
According to Newton's equations of motion (9-43), and the definition
(9-29), the first term in Eq. (9-46) is
V^ ™ (* dXi j- r, d y* _i_ -4 Bz <\ — V 1 (p dXi MF d ^ i a- w bz *\
f{ \ dq k dq k dq k / ££ \ dq k " dq k dq k /
= Qu. (9-47)
The derivatives appearing in the last term in Eq. (9-46) are calculated
as follows:
d dXi _ -1^ d 2 Xi . d 2 Xj _ _d_ /^ dx^ . . dxA _ dii
dtdq k ~ £{ dqkdqi Ql + dq k dt ~ dq k [^ dq t Ql + ~>t J ~ dq k '
(9-48)
where we have made use of Eq. (9-6). Similar expressions hold for y
and z. Thus the last sum in Eq. (9-46) is
V* i • — ^i j- • — ^Mi a- • — d g< i
£, m t \Xi j ( ^ + y t dt d ^ + Zi dt dq J
El . dXi . . dyi . . dZ{\ d -r^ 1 /-2 , .2 i .2\
.. . mi \ Xi eq- k + yi i- k + Zi WJ = W^ l hmi{Xi + yi+ Zi)
(9-4)
We have finally:
dT_
dqk
f = ^ + S' *=1.-".^- (9-50)
9-2] lagrange's equations 367
Our original expectation was not quite correct, in that we must add to
the generalized force Q k another term dT/dq k in order to get the rate of
change of momentum j> k - To see its meaning, consider the kinetic energy
of a particle in terms of plane polar coordinates, as given by Eq. (9-14).
In this case,
-gj = mre 2 , (9-51)
and if we make use of Eqs. (9-22) and (9-35), the equation of motion
(9-50) for qk = r is
mf = F r + mr0 2 . (9-52)
If we compare this with Eq. (3-207), which results from a direct applica-
tion of Newton's law of motion, we see that the term dT/dr is part of
the mass times acceleration which appears here transposed to the right
side of the equation. In fact, dT/dr is the "centrifugal force" which
must be added in order to write the equation of motion for r in the form
of Newton's equation for motion in a straight line. Had we been a bit more
clever originally, we should have expected that some such term might
have to be included. We may call dT/dqk a "fictitious force" which ap-
pears if the kinetic energy depends on the coordinate q k . This will be
the case when the coordinate system involves "curved" coordinates, that
is, if constant generalized velocities fa, ... , fan result in curved motions
of some parts of the mechanical system. Equations (9-50) are usually
written in the form
i(H)-ir«* * = i."-.3*-: (9-53)
If a potential energy exists, so that the forces Qk are derivable from
a potential energy function [Eq. (9-33)], we may introduce the Lagrangian
function
L(Qi, ..., Qsn; fa, ■■■ , fan) t) = T — V, (9-54)
where T depends on both q u . . . , q 3N and fa, ... , q 3N , but V depends
only on q i} . . . , q 3 jf (and possibly t), so that
A <H± = A <>T.
dt dfa dt dfa
(9-55)
dL dT dV dT .....
^ — = 5 n — = ^ — + Qk- (9-56)
dq k dq k dq k dq k
Hence Eqs. (9-53) can be written in this case in the form
a(i)-t = °- *-'.-.»• mi
368 lagrange's equations [chap. 9
In nearly all cases of interest in physics (although not in engineering),
the equations of motion can be written in the form (9-57). The most
important exception is the case where frictional forces are involved, but
such forces do not usually appear in atomic or astronomical problems.
Since Lagrange's equations have been derived from Newton's equations
of motion, they do not represent a new physical theory, but merely a differ-
ent but equivalent way of expressing the same laws of motion. As the
example of Eqs. (9-52) and (3-207) illustrates, the equations we get by
Lagrange's method can also be obtained by a direct application of Newton's
law of motion. However, in complicated cases it is usually easier to work
out the kinetic energy and the forces or potential energy in generalized
coordinates, and write the equations in Lagrangian form. Particularly in
problems involving constraints, as we shall see in Section 9-4, the La-
grangian method is much easier to apply. The chief value of Lagrange's
equations is, however, probably a theoretical one. From the manner in
which they were derived, it is evident that Lagrange's equations (9-57) or
(9-53) hold in the same form in any system of generalized coordinates. It
can also be verified by direct computation (see Problem 24) that if Eqs.
(9-57) hold in any coordinate system for any function L(g 1 , . . . , qzN',Qi,
■ ■ • , izN) t), then equations of the same form hold in any other coordinate
system. The Lagrangian function L has the same value, for any given set
of positions and velocities of the particles, no matter in what coordinate
system it may be expressed, but the form of the function L may be different
in different coordinate systems. The fact that Lagrange's equations have
the same form in all coordinate systems is largely responsible for their
theoretical importance. Lagrange's equations represent a uniform way of
writing the equations of motion of a system, which is independent of the
kind of coordinate system used. They form a starting point for more ad-
vanced formulations of mechanics. In developing the general theory of
relativity, in which cartesian coordinates may not even exist, Lagrange's
equations are particularly important.
9-3 Examples. We first consider a system of particles m 1; . . . , mx,
located by cartesian coordinates, and show that in this case Lagrange's
equations become the Newtonian equations of motion. The kinetic en-
ergy is
N
T = Y, i m ^ + y 2 i + if), (9-58)
and
*L = *L = *1 = o (9-59)
dxt dy { dZi '
Wi = m iXi , — = mm , — = miZi . (9-60)
9-4] SYSTEMS SUBJECT TO CONSTRAINTS 369
The generalized force associated with each cartesian coordinate is just
the ordinary force, as we see either from Eq. (9-29), or by comparing
Eq. (9-28) with Eq. (9-26). Hence the equations of motion (9-53) are
d(ar\
dt \dXi/
dt KdyJ dyi
a (mi\ _ *i_
dt \dzi) dzi
0L— •• - p
n — TfliXi — r ix }
dT = m iyi = F iv , [i = 1, . . . , N] (9-61)
rriiZi = F iz .
For a particle moving in a plane, the kinetic energy in polar coordi-
nates is given by Eq. (9-14), and the forces Q r and Q e by Eqs. (9-35).
The Lagrange equations are
mr — mrti 2 = F r> (9-62)
± ( mr 2 6) = rF e . (9-63)
These equations were obtained in Section 3-13 by elementary methods.
We now consider the rotating coordinate system denned by Eqs. (9-4)
or (9-5). The kinetic energy is given by Eq. (9-16), and the generalized
forces Q r and Q t will be the same as in the previous example. Lagrange's
equations in this case are
mr — mrt) 2 — 2rruor6 — mo) 2 r = F r , (9-64)
j (mr 2 d) + 2mwrf = rF e . (9-65)
The reader should verify that the third term on the left in Eq. (9-64)
is the negative of the coriolis force in the r-direction due to the rotation
of the coordinate system, and that the fourth term is the negative of
the centrifugal force. The second term in Eq. (9-65) is the negative of
the coriolis torque in the 0-direction. Thus the necessary fictitious forces
are automatically included when we write Lagrange's equations in a
moving coordinate system. It must be noticed, however, that we use the
actual kinetic energy [Eq. (9-16)] with respect to a coordinate system
at rest, expressed in terms of the rotating coordinates, and not the kinetic
energy as it would appear in the rotating system if we ignored the motion
of the coordinate system.
9-4 Systems subject to constraints. One important class of mechanical
problems in which Lagrange's equations are particularly useful comprises
systems which are subject to constraints.
370 laghange's equations [chap. 9
A rigid body is a good example of a system of particles subject to
constraints. A constraint is a restriction on the freedom of motion of a
system of particles in the form of a condition which must be satisfied by
their coordinates, or by the allowed changes in their coordinates. For
example, a very simple hypothetical rigid body would be a pair of par-
ticles connected by a rigid weightless rod of length I. These particles are
subject to a constraint which requires that they remain a distance I apart.
In terms of their cartesian coordinates, the constraint is
[(* 2 - *i) 2 + (2/2 - 2/i) 2 + (*a - zi) 2 ] 1/2 = 1 (9-66)
If we use the coordinates X, Y, Z of the center of mass and spherical
coordinates r, 6, <p to locate particle 2 with respect to particle 1 as origin,
the constraint takes the simple form:
r=l. (9-67)
There are thus only five coordinates X, Y, Z, 6, <p left to determine. Each
constraint which can be expressed in the form of an equation like (9-66)
enables us to eliminate one of the coordinates by choosing coordinates in
such a manner that one of them is held constant by the constraint. For a
rigid body, the constraints require that the mutual distances of all pairs
of particles remain constant. For a body containing N particles, there
are %N(N — 1) pairs of particles. However, it is not hard to show that
it is sufficient to specify the mutual distances of 3N — 6 pairs, UN > 3.
Hence we can replace the 3N cartesian coordinates of the N particles by
3iV — 6 mutual distances, 3 coordinates of the center of mass, and 3 coor-
dinates describing the orientation of the body. Since the 3N — 6 mutual
distances are all constant, the problem is reduced to one of finding the
motion in terms of six coordinates. Another example of a system subject
to a constraint is that of a bead sliding on a wire. The wire is situated
along a certain curve in space, and the constraints require that the posi-
tion of the bead lie on this curve. Since the coordinates of the points
along a space curve satisfy two equations (e.g., the equations of two
surfaces which intersect along the curve), there are two constraints, and
we can locate the position of the bead by a single coordinate. (Can you
suggest a suitable coordinate?) If the wire is moving, we have a moving
constraint, and our single coordinate is relative to a moving system of
reference. Constraints which can be expressed in the form of an equa-
tion relating the coordinates are called holonomic. All the above examples
involve holonomic constraints.
Constraints may also be specified by a restriction on the velocities,
rather than on the coordinates. For example, a cylinder of radius a,
rolling and sliding down an inclined plane, with its axis always horizontal,
9-4]
SYSTEMS SUBJECT TO CONSTRAINTS
371
Fig. 9-3. A cylinder rolling down Fig. 9-4. A disk rolling on a hori-
an incline. zontal plane.
can be located by two coordinates s and 0, as in Fig. 9-3. The coordi-
nate s measures the distance the cylinder has moved down the plane,
and the coordinate 9 is the angle that a fixed radius in the cylinder has
rotated from the radius to the point of contact with the plane. Now
suppose that the cylinder is rolling without slipping. Then the velocities
s and 6 must be related by the equation
s = a&,
which may also be written
ds = a d$.
This equation can be integrated :
— ad
C,
(9-68)
(9-69)
(9-70)
where C is a constant. This equation is of the same type as Eq. (9-66),
and shows that the constraint is holonomic, although it was initially ex-
pressed in terms of velocities. If a constraint on the velocities, like Eq.
(9-68), can be integrated to give a relation between the coordinates, like
Eq. (9-70), then the constraint is holonomic. There are systems, however,
in which such equations of constraint cannot be integrated. An example
is a disk of radius a rolling on a horizontal table, as in Fig. 9-4. For sim-
plicity, we assume that the disk cannot tip over, and that the diameter
which touches the table is always vertical. Four coordinates are required
to specify the position of the disk. The coordinates x and y locate the
point of contact on the plane; the angle <p determines the orientation of
the plane of the disk relative to the z-axis; and the angle 6 is the angle
between a radius fixed in the disk and the vertical. If we now require
that the disk roll without slipping (it can also rotate about the vertical
axis), this implies two equations of constraint. The velocity of the point
of contact perpendicular to the plane of the disk must be zero:
x sin <p + y cos tp = 0,
(9-71)
372 lagrange's equations [chap. 9
and the velocity parallel to the plane of the disk must be
x cos <p — y sin <p = ad. (9-72)
It is not possible to integrate these equations to get two relations between
the coordinates x, y, 8, <p. To see this, we note that by rolling the disk
without slipping, and by rotating it about a vertical axis, we can bring
the disk to any point x, y, with any angle <p between the plane of the disk
and the x-axis, and with any point on the circumference of the disk in
contact with the table, i.e., any angle 0. For if the disk is at any point
x, y, and the desired point on the circumference is not in contact with the
table, we may roll the disk around a circle whose circumference is of
proper length, so that when it returns to x, y, the desired point will be in
contact with the table. It may then be rotated to the desired angle <p.
This shows that the four coordinates x, y, 6, <p are independent of one
another, and there cannot be any relation between them. It must there-
fore be impossible to integrate Eqs. (9-71) and (9-72), and consequently
this is an example of a nonholonomic constraint.
The number of independent ways in which a mechanical system can
move without violating any constraints which may be imposed is called
the number of degrees of freedom of the system. To be more precise, the
number of degrees of freedom is the number of quantities which must be
specified in order to determine the velocities of all particles in the system
for any motion which does not violate the constraints. For example, a
single particle moving in space has three degrees of freedom, but if it is
constrained to move along a certain curve, it has only one. A system
of N free particles has 3N degrees of freedom, a rigid body has 6 degrees
of freedom (three translational and three rotational), and a rigid body
constrained to rotate about an axis has one degree of freedom. The
disk shown in Fig. 9-4 has four degrees of freedom if it is allowed to
slip on the table, because we need then to specify x, y, 0, <p. But if the
disk is required to roll without slipping, there are only two degrees of
freedom, because if <i> and any one of the velocities x, y, 6 are given, the
remaining two can be found from Eqs. (9-71) and (9-72). The disk is
only free to roll, and to rotate about a vertical axis. For holonomic sys-
tems, the number of degrees of freedom is equal to the minimum number
of coordinates required to specify the configuration of the system when
coordinates held constant by the constraints are eliminated. Nonholo-
nomic constraints occur in some problems in which bodies roll without
slipping, but they are not of very great importance in physics. We shall
therefore restrict our attention to holonomic systems.
For a holonomic system of N particles subject to c independent con-
straints, we can express the constraints as c relations which must hold
9-4] SYSTEMS SUBJECT TO CONSTRAINTS 373
between the 3iV cartesian coordinates (including possibly the time if the
constraints are changing with time) :
hi(xi, 2/i, ... , zn) t) = a x ,
hzfri, 2/i, ■ ■ • , zn) t) = a 2 , (9 _ ?3 .
h c {xi, 2/i, . . . ,z N ;t) = a c ,
where hi, . . . , h c are c specified f mictions. The number of degrees of
freedom will be
/ = 3JV - c. (9-74)
As Eqs. (9-73) are independent, we may solve them for c of the SN car-
tesian coordinates in terms of the other 3N — c coordinates and the
constants oi, . . . , a c . Thus only ZN — c coordinates need be specified,
and the remainder can be found from Eqs. (9-73) if the constants
ai, . . . , a c are known. We may take as generalized coordinates these
3iV" — c cartesian coordinates and the c quantities oi, . . . , a c defined by
Eqs. (9-73), and held constant by the constraints. Or we may define
3N — c generalized coordinates qi, . . . , q/, in any convenient way:
?i = ?i(*i, 2/i, • • • , zn] t),
q 2 = q 2 (xi, J/i, . . . ,z N ;t), , g
Qf = Qf(xi, 2/i, ... , zn; t).
Equations (9-73) and (9-75) define a set of 3iV coordinates qi, . . . , q/;
ai, . . . , a c , and are analogous to Eqs. (9-1). They may be solved for the
cartesian coordinates:
(9-76)
Now let Qi, . . . , Qf, Qf + i, . . . , Qf+ C be the generalized forces corre-
sponding to the coordinates qi, . . . , gy; Oi, . . . , a c . We have then a set
of Lagrange equations for the constrained coordinates and another for
the unconstrained coordinates:
,/, (9-77)
..,c;c+f=3N. (9-78)
*\ = xi(qi, ■ •
• . Qf, Oi, • •
. , a c ; t),
2/i = 2/i(«?i, • ■
• , Q/; 0,1, . .
. , a c ; t),
zn — ZN(qi, ■ •
. , q f ; a lt . .
• , a c ; t).
d dT
dt dqk
dT -O
-dq~ k - Qk '
k = 1, .
d dT
dt ddj
dT a
3= h
374 lagrange's equations [chap. 9
The importance of this separation of the problem into two groups of
equations is that the forces of constraint can be so chosen that they do
no work unless the constraints are violated, as we shall show in the next
paragraph. If this is true, then according to the definition (9-30) of the
generalized force, the forces of constraint do not contribute to the gen-
eralized force Qk associated with an unconstrained coordinate q^. Since
the values of the constrained coordinates a\, . . . , a c are held constant,
we can solve Eqs. (9-77) for the motion of the system in terms of the
coordinates g 1( . . . , q/, treating oi, . . . , a c as given constants, without
knowing the forces of constraint. This is a great advantage, for the
forces of constraint depend upon how the system is moving, and cannot,
in general, be determined until after the motion has been found. All
we usually know about the constraining forces is that they have whatever
values are required to maintain the constraints. Having solved Eqs. (9-77)
for q x (<),..., q/(t), we may then, if we wish, substitute these functions
in Eqs. (9-78) and calculate the forces of constraint. This may be a
matter of considerable interest to the engineer who needs to verify that
the constraining members are strong enough to withstand the constrain-
ing forces. Lagrange's equations thus reduce the problem of finding the
motion of any holonomic system with / degrees of freedom to the problem
of solving / second-order differential equations (9-77). When we speak
of the generalized coordinates, the constrained coordinates ai, . . . , a c
may or may not be included, as convenient.
If a bead slides on a frictionless wire, the wire can only exert constrain-
ing forces perpendicular to itself, so that no work is done on the bead
so long as it stays on the wire.* If there is friction, we can separate the
force on the bead into a component perpendicular to the wire which
holds the bead on the wire without doing any work, and a frictional
component along the wire which does work and will therefore have to be
included in the generalized force associated with motion along the wire.
If the. frictional component depends on the perpendicular component, as
it does for dry sliding friction, then we cannot solve Eqs. (9-77) first,
independently of Eqs. (9-78), and one great advantage of the Lagrangian
method is lost. If two particles are held a fixed distance apart by a rigid
rod, then by Newton's third law, the force exerted by the rod on one
particle is equal and opposite to that on the other. It was shown in
* If the wire is moving, the force exerted by the wire may do work on the
bead, but the virtual displacements in terms of which the generalized forces
have been defined are to be imagined as taking place at a fixed instant of time,
and for such a displacement which does not violate the constraints, no work
is done. Hence even in the case of moving constraints, the constraining forces
do not appear in the generalized forces associated with the unconstrained
coordinates.
9-5]
EXAMPLES OF SYSTEMS SUBJECT TO CONSTRAINTS
375
Section 5-1 that no net work is done on the system by the rod so long as
the constraint is not violated, that is, so long as the rod is not stretched
or compressed. A similar situation will be found in all other cases; the
constraints could always be maintained by forces which do no work.
If the forces Qi, . . . , Qf are derivable from a potential energy func-
tion, then we can define a Lagrangian function L(q l; . . . , q f ; q u . . . , q/)
which may in some cases depend on t, and which may also depend on the
constants a\, . . . , a c . The first / Lagrange equations (9-77) can then
be written in the form
A ik _ Ok = o
dt dq k dq k ~ '
h= 1, ...,/.
(9-79)
9-5 Examples of systems subject to constraints. A simple mechanical
system involving constraints is the Atwood's machine shown in Fig. 9-5.
Weights mj, jre 2 are connected by a rope of length I over a fixed pulley.
We assume the weights move only vertically, so that we have only one
degree of freedom. We take as coordinates the distance x of m\ below
the pulley axle, and I, the length of the rope. The coordinate I is con-
strained to have a constant value, and could be left out of consideration
from the start if we wish only to find the motion. If we also want to
find the tension in the string, we must include I as a coordinate. The
kinetic energy is
T = %m x x 2 + %m 2 (i - x) 2 . (9-80)
The only forces acting on mi and m 2 are the tension t in the rope and the
force of gravity. The work done when x increases by Sx, I remaining
\m,2
Fig. 9-5. Atwood's machine.
376 lagrange's equations [chap. 9
constant, is
SW = (mig — t) 8x — (m 2 g — r) 8x
= (mi — m 2 )g Sx = Q x Sx, (9-81)
so that
Q x = (mi - m a )g. (9-82)
Note that Q x is independent of t. The work done when I increases by
81, x remaining constant, is
SW = {m 2 g - t) 81 = Qi 81, (9-83)
so that
Qi = m 2 g - t. (9-84)
Notice that in order to obtain an equation involving the force of con-
straint r, we must consider a motion which violates the constraint. This
is also true if we wish to measure a force physically; we must allow at
least a small motion in the direction of the force. The Lagrange equations
of motion are (since I = l = 0)
i\w~^ = (mi + m2)£ = (mi _ mz)9 > (9_85)
d (dT\ dT .. , n ...
It \a) ~ aT = ~ m2X = m * g ~ T - (9 " 86)
The first equation is to be solved to find the motion:
x = Xo + Vot + $ Wl ~^ 2 gt\ (9-87)
mi -f- m 2
The second equation can then be used to find the tension t necessary to
maintain the constraint :
, . .., 2mim 2 , ,
r = m 2 (g + x) = ^^^ g. (9-88)
In this case the tension is independent of time and can be found from
Eqs. (9-85) and (9-86) immediately, although in most cases the con-
straining forces depend on the motion and can be determined only after
the motion is found. Equations (9-85) and (9-86) have an obvious phys-
ical interpretation and could be written down immediately from elemen-
tary considerations, as was done in Section 1-7.
A problem of little practical importance, but which is quite instructive,
is that in which one cylinder rolls upon another, as shown in Fig. 9-6.
The cylinder of radius a is fixed, and the cylinder of radius aa rolls around
it under the action of gravity. Suppose we are given that the coefficient
9-5]
EXAMPLES OF SYSTEMS SUBJECT TO CONSTRAINTS
377
Fig. 9-6. One cylinder rolling on another.
of static friction between the cylinders is n, the coefficient of sliding
friction is zero,* and that the moving cylinder starts from rest with its
center vertically above the center of the fixed cylinder. We shall assume
that the axis of the moving cylinder remains horizontal during the motion.
It is advisable in all problems, and essential in this one, to think carefully
about the motion before attempting to find the mathematical solution.
It is clear that the moving cylinder cannot roll all the way around the
fixed cylinder, for the normal force F which is exerted by the fixed cyl-
inder on the moving one can only be directed outward, never inward.
Therefore at some point, the moving cylinder will fly off the fixed one.
The point at which it flies off is the point at which
F = 0.
(9-89)
Furthermore, the cylinder cannot continue to roll without slipping right
up to the point at which it flies off, for the frictional force f which pre-
vents slipping is limited by the condition
f<tf,
(9-90)
and will certainly become too small to prevent slipping before the point
at which Eq. (9-89) holds. The motion therefore is divided into three
parts. At first the cylinder rolls without slipping through an angle t
determined by the condition
/ = nF. (9-91)
* This implies that the moving cylinder either rolls without slipping, if the
static friction is great enough, or slips without any friction at all. The latter
assumption is made to simplify the problem.
378 lagrange's equations [chap. 9
Beyond the angle 0i, the cylinder slides without friction until it reaches
the angle 2 determined .Jby Eq. (9-89), after which it leaves the fixed
cylinder and falls freely. We may anticipate some mathematical difficulties
with the initial part of the motion due to the fact that the initial posi-
tion of the moving cylinder is one of unstable equilibrium. Physically
there is no difficulty, since the slightest disturbance will cause the cylinder
to roll down, but mathematically there may be a difficulty which we must
watch out for, inasmuch as the needed slight disturbance will not appear
in the equations.
Let us find that part of the motion when the moving cylinder rolls with-
out slipping. There is then only one degree of freedom, and we shall
specify the position of the cylinder by the angle between the vertical and
the line connecting the centers of the two cylinders. In order to compute
the kinetic energy, we introduce the auxiliary angle <p through which the
moving cylinder has rotated about its axis. The condition that the cylin-
der roll without slipping leads to the equation of constraint:
ad = aa(<p — 0), (9-92)
which can be integrated in the form
(1 + a)6 = a<p. (9-93)
If we were concerned only with the rolling motion, we could now proceed
to set up the Lagrange equation for 6, but inasmuch as we need to know
the forces of constraint F and f, it is necessary to introduce additional co-
ordinates which are maintained constant by these constraining forces.
The frictional force f maintains the constraint (9-93), and an appropriate
coordinate is
7=0- -25L. . (9-94)
So long as the cylinder rolls without slipping, 7 = 0; 7 measures the angle
of slip around the fixed cylinder. The normal force F maintains the dis-
tance r between the centers of the cylinders:
r = a + aa= (1 + a)a. (9-95)
The kinetic energy of the rolling cylinder is the energy associated with the
motion of its center of mass plus the rotational energy about the center of
mass:
T = %m(r 2 + rH 2 ) + %I<p 2 . (9-96)
After substituting <p from Eq. (9-94), and since I = %ma 2 a 2 , for a solid
cylinder of radius aa, we have
T = imr 2 + imr 2 6 2 + |m(l + a) 2 a 2 (6 2 - 270 + 7 2 ). (9-97)
9-5] EXAMPLES OF SYSTEMS SUBJECT TO CONSTRAINTS 379
The equations of constraint [Eq. (9-95) and 7=0] must not be used until
after the equations of motion are written down. The generalized forces
are most easily determined with the help of Eq. (9-30); they are*
Q, = mgr sin 6, (9-98)
Qy = -fad + ex), (9-99)
Q r = F — mg cos 0. (9-100)
The Lagrange equations for 6, 7, and r are now
m[r 2 + \a?(l + a) 2 ]S + 2mrf6 — - Jma 2 (l + a) 2 7 = mgr sin 6, (9-101)
-Jma 2 (l + a) 2 S + ima 2 (l + a) 2 y = ~fa(l + a), (9-102)
mr — mrd 2 = F — mg cos 0. (9-103)
We can now insert the constraints 7=0 and r = (1 + a)a, so that these
equations become
f (1 + a) 2 ma 2 = (1 + a)mga sin d, (9-104)
/ = i(l + a)ma§, (9-105)
F = mg cosd — (1 + a)ma6 2 . (9-106)
Had we ignored the terms involving y in the kinetic energy, the 6 equation,
which determines the motion, would have come out correctly, but the
equation for the constraining force f would have been missing a term.
This happens when the constrained coordinates are not orthogonal to the
unconstrained coordinates, since a cross term (70) then appears in the
kinetic energy.
The equation of motion (9-104) can be solved by the energy method.
The total energy, so long as the cylinder rolls without slipping, is
4(1 + a) 2 maH 2 + (1 + a)mga cos B = E, (9-107)
and is constant, as can easily be shown from Eq. (9-104), and as we
know anyway since the gravitational force is conservative and the forces
of constraint do no work. Since the moving cylinder starts from rest
at 6 = 0,
E = (1 + a)mga. (9-108)
! The reader will find it an instructive exercise to verify these formulas.
380 lagrange's equations [chap. 9
We substitute this in Eq. (9-107) and solve for 0:
6 = 2 (^ J'' 2 sin | , (9-109)
where
^mr^y (9 - 110)
We can now integrate to find 6(t):
f A* = (*L) 1,a f &, ( 9-iii)
Jo sin 6/2 \a/ Jo ' v
[ toto ll -(?)""'■ (9 - 112)
When we substitute the lower limit = 0, we run into a difficulty, for
In = — oo ! This is the expected difficulty due to the fact that =
is a point of equilibrium, albeit unstable. If there is no disturbance what-
ever, it will take an infinite time for the cylinder to roll off the equilibrium
point. Let us suppose, however, that it does roll off due to some slight
disturbance, and let us take the time t = as the time when the angle
has some small value O . There is now no difficulty, and we have
tan | = (tan ^) exp [(f) 1 '' *] ■ (9-113)
As t — > oo, — > 2ir, and the moving cylinder rolls all the way around
the fixed one, if the constraints continue to hold. The rolling constraint
holds, however, only so long as Eq. (9-90) holds. When we substitute
from Eqs. (9-105), (9-106), and (9-109), Eq. (9-90) becomes
%mg sin < ifimg(7 cos — 4). (9-114)
At 6 = 0, this certainly holds, so that the cylinder does initially roll,
as we have supposed. At 8 = 7r/2, however, it certainly does not hold,
since the left member is then positive and the right, negative. The angle 0i
at which slipping begins is determined by the equation
sin 0i = n(7 cos 0i — 4), (9-115)
whose solution is
™=a 28m 2 + [1 + 33 M 2 ] 1/2 • , Q 11ft ,
C0S 01 = 1 + 49m 2 ( }
9-6] CONSTANTS OF THE MOTION AND IGNOBABLE COOBDINATES 381
The second part of the motion, during which the moving cylinder slides
without friction around the fixed one, can be found by solving Eqs. (9-101)
and (9-102) for 6(t), 7(0, with / = and with only the single constraint
r = (1 + <x)a, and with initial values 6 = 6 U = 1; determined from
Eqs. (9-116) and (9-109). The solution can be found without essential
difficulty, and the angle 2 at which the moving cylinder leaves the fixed
one can then be determined from Eqs. (9-106) and (9-89). These calcu-
lations are left to the reader.
9-6 Constants of the motion and ignorable coordinates. We remarked
in Chapter 3 that one general method for solving dynamical problems is
to look for constants of the motion, that is, functions of the coordinates
and velocities which are constant in time. One common case in which
such constants can be found arises when the dynamical system is charac-
terized by a Lagrangian function in which some coordinate ft does not
occur explicitly. The corresponding Lagrange equation (9-57) then re-
duces to
d
dt \dft>
(i)=°- <•"»")
This equation can be integrated immediately:
T~r- = Pk = a constant. (9-118)
"ft
Thus, whenever a coordinate ft does not occur explicitly in the Lagrangian
function, the corresponding momentum pu is a constant of the motion.
Such a coordinate ft is said to be ignorable. If ft is ignorable, we can
solve Eq. (9-118) for ft in terms of the other coordinates and velocities,
and of the constant momentum p k , and substitute in the remaining
Lagrange equations to eliminate ft and reduce by one the number of vari-
ables in the problem; (ft was already missing from the equations, since
it was assumed ignorable.) When the remaining variables have been
found, they can be substituted in Eq. (9-118), to give ft as a function
of t; ft is then obtained by integration. If all but one of the coordinates
are ignorable, the problem can thus be reduced to a one-dimensional
problem and solved by the energy integral method, if L does not depend
on the time t explicitly.
For example, in the case of central forces, the potential energy depends
only on the distance r from the origin, so that if we use polar coordinates
r, in a plane, V is independent of 0. Since T is also independent of
according to Eq. (9-14) (T depends of course on 6), we will have
fe = h < r - y ) = °« < 9 - 119 >
382 lagrange's equations [chap. 9
and hence
= mr 6 = p 9 = a constant, (9-120)
86
a result which we obtained in Section 3-13 by a different argument. We
see that the constancy of p e is a result of the fact that the system is sym-
metrical about the origin, so that L cannot depend on 8. If a system of
particles is acted on by no external forces, then if we displace the whole
system in any direction, without changing the velocities and relative
positions of the particles, there will be no change in T or V, or in L. If
X, Y, and Z are rectangular coordinates of the center of mass, and if
the remaining coordinates are relative to the center of mass, so that
changing X corresponds to displacing the whole system, then
H = 0, (9-121)
and therefore px, the total linear momentum in the x-direction, will be
constant, a result we proved in Section 4-1 by a different method.
It is of interest to see how to show from Lagrange's equations that the
total energy is a constant of the motion. In order to find an energy inte-
gral of the equations of motion in Lagrangian form, it is necessary to
know how to express the total energy in terms of the Lagrangian func-
tion L. To this end, let us consider a system described in terms of a fixed
system of coordinates, so that the kinetic energy T is a homogeneous
quadratic function of the generalized velocities <ji, . . . , <j/ [i.e., T x =
T = in Eq. (9-13)]. By Euler's theorem,* we have
t & g = 2f. (9-122)
k=l * K
Thus if
L = T 2 - V, (9-123)
where V is a function of the coordinates q lt . . . , q/ alone, then, by Eq.
(9-122),
1 £i h £-L=T+V = E. (9-124)
We now consider the time derivative of the left member of Eq. (9-124).
For greater generality, we shall at first allow L to depend explicitly on t.
* W. F. Osgood, Advanced Calculus. New York: Macmillan, 1937. (Page 121.)
The reader unfamiliar with Euler's theorem can readily verify Eq. (9-122) for
himself by substituting for T = T% from Eq. (9-9).
9-6] CONSTANTS OF THE MOTION AND IGNOKABLE COORDINATES 383
In the case we have considered, L does not depend explicitly on t. There
are cases, however, when a system is subject to external forces that change
with time and that can be derived from a potential V that varies with
time. An example would be an atom subject to a varying external electric
field. In such cases, the equations of motion can be written in the La-
grangian form (9-57) with the Lagrangian depending explicitly on the
time t. In the case of moving coordinate systems also, the Lagrangian
may depend on the time even though the forces are conservative. The
time derivative of the left member of Eq. (9-124) is
d ( <A . dL T \ <AL dL . . d/dL\ dL . dL .. 1 dL
^A . \d (dL\ dL\ dL dL , n 10rN
If L does not depend explicitly on t, the right side of Eq. (9-125) is zero,
and
^2 § k ~p L = & constant. (9-126)
When L has the form (T 2 — V), as in a stationary coordinate system,
this is the conservation of energy theorem. Regardless of the form of L,
Eq. (9-126) represents an integral of Lagrange's equations (9-57), when-
ever L does not contain t explicitly, but the constant quantity on the
left is not always the total energy. Note the analogy between the con-
servation of generalized momentum p k when L is independent of qt, and
the conservation of energy when L is independent of t. There are many
ways in which the relation between time and energy is analogous to the
relation between a coordinate and the corresponding momentum.
We have seen that the familiar conservation laws of energy, momentum,
and angular momentum can be regarded as consequences of symmetries
exhibited by the mechanical systems to which they apply; that is, they are
consequences of the fact that the Lagrangian function L, which determines
the equations of motion, is independent of time and of the position and
orientation of the entire system in space. This result, derived here for
classical mechanics, holds generally throughout physics. In quantum
mechanics and in relativity theory, even when we include electromagnetic
and other kinds of force fields, conservation laws are associated with sym-
metries in the fundamental equations. We might, for example, define
energy as that quantity which is constant because the laws of physics are
always the same (if indeed they are!).
384 lagrange's equations [chap. 9
9-7 Further examples. The spherical pendulum is a simple pendulum
free to swing through the entire solid angle about a point. The pendulum
bob is constrained to move on a spherical surface of radius R. We locate
the bob by the spherical coordinates 8, <p (Fig. 9-7). We may include
the length R of the pendulum as a coordinate if we wish to find the ten-
sion in the string, but we omit it here, as we are concerned only with
finding the motion. If the bob swings above the horizontal, we will sup-
pose that it still remains on the sphere, which would be true if the string
were replaced by a rigid rod. Otherwise the constraint disappears when-
ever a compressional stress is required to maintain it, since a string will
support only a tension and not a compression. The velocity of the bob is
v = Rd\ + R sin 8 <pm. (9-127)
Hence the kinetic energy is
T = \mv 2 = %mR 2 8 2 + %mR 2 sin 2 <p 2 . (9-128)
The potential energy due to gravity, relative to the horizontal plane, is
V = mgR cos 0. (9-129)
Hence the Lagrangian function is
L = T — V = %mR 2 8 2 + %mR 2 sin 2 <p 2 — mgR cos 0. (9-130)
The Lagrange equations are
j
(mR 2 6) — mR 2 <p 2 sin 6 cos 6 — mgR sin 8=0, (9-131)
dt
dt
^ (mR 2 sm 2 d<p) = 0. (9-132)
The coordinate <p is ignorable, and the second equation can be integrated
immediately:
mR 2 sin 2 <p = p v = a constant. (9-133)
Also, since
^ = 0, (9-134)
the quantity
*?* + *?£•-■&= %mR 2 6 2 + %mR 2 sin 2 <p 2 + mgR cos 8 (9-135)
60 o<p
is constant, by Eq. (9-126). We recognize the quantity on the right as
the total energy, as it should be, since we are using a fixed coordinate
9-7]
FURTHER EXAMPLES
385
MgR
-MgR
Fig. 9-7. A spherical pendulum.
Fig. 9-8. Effective potential 'V'(fi)
for spherical pendulum.
system. Calling this constant E, and substituting for <p from Eq. (9-133),
we have
vl
hnR 2 r + J*. 2Q +. mgR cos = E.
2m R 2 sin 2 6
We may introduce an effective potential 'V'(6) for the motion:
„2
'V'(d) = mg R cos +
so that
2mR 2 sin 2
%mR 2 2 = E — l V'(d).
(9-136)
(9-137)
(9-138)
Since the left member cannot be negative, the motion is confined to those
values of for which 'V'(6) < E. The effective potential 'V'(O) is plotted
in Fig. 9-8. We see that for p v = 0, 'V'(0) is the potential curve for a
simple pendulum, with a minimum at = tt and a maximum at = 0.
For E = —mgR, the pendulum is at rest at 6 = ir. For mgR > E >
—mgR, the pendulum oscillates about = ir. For E > mgR, the pendulum
swings in a circular motion through the top and bottom points 6 =
and ir. When p v 5* 0, the motion is no longer that of a simple pendulum,
and 'V'(6) now has a minimum at a point 6 between tc/2 and ir, and
rises to infinity at = and = ir. The larger p v , the larger the mini-
mum value of 'F(0), and the closer O is to w/2. If E = 'V'(6 ), then
is constant and equal to O , and the pendulum swings in a circle about
the vertical axis. As p r — > oo, the pendulum swings more and more
nearly in a horizontal plane. For E > 'V'(0 O ), oscillates between a
maximum and minimum value while the pendulum swings about the
vertical axis. The reader should compare these results with his mechan-
ical intuitions or his experience regarding the motion of a spherical
386 lagranue's equations [chap. 9
pendulum. The solution of Eq. (9-138) for 0(0 cannot be carried out
in terms of elementary functions, but we can treat circular and nearly
circular motions very easily. The relation between p f and O for uniform
circular motion of the pendulum about the z-axis is
[d'V'~\ D ■ a vl cos O , „ Q v
— — = —mgR sin O 1 . , „ = 0. (9-139)
L dd ie v mR 2 sin 8 O
It is evident from this equation that O > ir/2, and that O — > tt/2 as
p v — > oo. By substituting from Eq. (9-133), we obtain a relation be-
tween tp and O for uniform circular motion:
Z? = .£ ?_^ . (9-140)
^ R (— cos0 o )
The energy for uniform circular motion at an angle O , if we use Eqs.
(9-136) and (9-139), and the fact that = 0, is
mgR / 2-3sin 2 0<A . }
2 \ cos O /
For an energy slightly larger than E , and an angular momentum p v
given by Eq. (9-139), the angle will perform simple harmonic oscilla-
tions about the value O . For if we set
k
\d 2 'V>
I d02 .
mgi? (1 + 3 cos 2 O ), (9-142)
Je —cos O
then, for small values of — O , we can expand 'V'{B) in a Taylor series:
'F(0) = E + Jfc(0 - O ) 2 . (9-143)
The energy equation (9-138) now becomes
%mRH 2 + 4fc(tf - O ) 2 = E - E Q . (9-144)
This is the energy for a harmonic oscillator with energy E — E , coordi-
nate — O) mass mR 2 , spring constant k. The frequency of oscillation
in is therefore given by
2 k g 1 + 3 cos 2 0p , s
W = m ~R? = R -cos0 o * (9 ~ M5)
This oscillation in is superposed upon a circular motion around the
z-axis with an angular velocity <p given by Eq. (9-133); <p will vary
slightly as 6 oscillates, but will remain very nearly equal to the constant
value given by Eq. (9-140). It is of interest to compare <j> and w:
9-7]
FURTHER EXAMPLES
387
■ 2
1
1 + 3 C0S2 6
(9-146)
Since 6 > t/2, this ratio is less than 1, so that co > <j>, and the pendulum
wobbles up and down as it goes around the circle. At O = tt/2, <p = w,
and the pendulum moves in a circle whose plane is tilted slightly from the
horizontal; this case occurs only in the limit of very large values of p v . It
is clear physically that when p v is so large that gravity may be neglected,
the motion can be a circle in any plane through the origin. Can you show
this mathematically? Near o = 0, w = 2<p, so that 6 oscillates twice per
revolution and the pendulum bob moves in an ellipse whose center is on
the z-axis. This corresponds to the motion of the two-dimensional har-
monic oscillator discussed in Section 3-10, with equal frequencies in the
two perpendicular directions.
As a last example, we consider a system in which there are moving
constraints. A bead of mass m slides without friction on a circular hoop
of radius a. The hoop lies in a vertical plane which is constrained to
rotate about a vertical diameter with constant angular velocity w. There
is just one degree of freedom, and inasmuch as we are not interested in
the forces of constraint, we choose a single coordinate 8 which measures
the angle around the circle from the bottom of the vertical diameter to
the bead (Fig. 9-9). The kinetic energy is then
T = ±ma 2 d 2 + frnaW sin 2 0,
and the potential energy is
V = —mga cos 6.
The Lagrangian function is
%ma 2 6~ 2 + Jmo 2 w 2 sin 2 6 + mga cos 0.
(9-147)
(9-148)
(9-149)
'1
/>
+mga-
t/2 y^ /
a) < a*,*
7T
— mga-
V*GJ > (tic
Fig. 9-9. A bead sliding on a rota- Fig. 9-10. Effective potential en-
ting hoop. ergy for system shown in Fig. 9-9.
388 lagrange's equations [chap. 9
The Lagrange equation of motion can easily be written out, but this is
unnecessary, for we notice that
at u '
and therefore, by Eq. (9-126), the quantity
0~- L= fynaH 2 - 4mo V sin 2 6 - mga cos = 'E' (9-150)
do
is constant. The constant 'E' is not the total energy T + V, for the
middle term has the wrong sign. The total energy is evidently not con-
stant in this case. (What force does the work which produces changes
in T + V?) We may note, however, that we can interpret Eq. (9-149)
as a Lagrangian function in terms of a fixed coordinate system with the
middle term regarded as part of an effective potential energy:
<y(0) = —^ma 2 u 2 s in 2 6 — mga cos 0. (9-151)
The energy according to this interpretation is 'E'. The first term in 'V'(d)
is the potential energy associated with the centrifugal force which must
be added if we regard the rotating system as fixed. The effective poten-
tial is plotted in Fig. 9-10. The shape of the potential curve depends on
whether w is greater or less than a critical angular velocity
w c = (g/a) 1 ' 2 . (9-152)
It is left to the reader to show this, and to discuss the nature of the motion
of the bead in the two cases.
9-8 Electromagnetic forces and velocity-dependent potentials. If the
forces acting on a dynamical system depend upon the velocities, it may
be possible to find a function U(qi, . . . , q/; qi, . . . , q/; t) such that
*-!*-«' s=1 '■ (M53)
If such a function U can be found, then we can define a Lagrangian
function
L= T - U, (9-154)
so that the equations of motion (9-53) can be written in the form (9-57) :
9-8] ELECTROMAGNETIC FORCES, VELOCITY- DEPENDENT POTENTIALS 389
The function U may be called a velocity-dependent potential. If there are
also forces derivable from an ordinary potential energy V(q u . . . , q/), V
may be included in U, since Eq. (9-153) reduces to Eq. (9-33) for those
terms which do not contain the velocities. The function U may depend
explicitly on the time t. If it does not, and if the coordinate system is a
fixed one, then L will be independent of t, and the quantity
E = Z & f| - L > (9-156)
will be a constant of the motion, according to Eq. (9-126). In this case,
we may say that the forces are conservative even though they depend on
the velocities. It is clear from this result that it cannot be possible to
express frictional forces in the form (9-153), for the total energy is not
constant when there is friction unless we include heat energy, and heat
energy cannot be defined in terms of the coordinates and velocities
Qu ■ ■ ■ > Qf', Qi, ■ ■ ■ , if, and hence cannot be included in Eq. (9-156).
It is not hard to show that if the velocity-dependent parts of U are linear
in the velocities, as they are in all important examples, the energy E
defined by Eq. (9-156) is just T + V, where V is the ordinary potential
energy and contains the terms in U that are independent of the velocities.
As an example, a particle of charge q subject to a constant magnetic
field B is acted on by a force (guassian units)
F = I v X B, (9-157)
c
or
F* = ? (SB. ~ ZB V ),
c
F v = l (zB x - *B.), (9-158)
c
F z = I (xB v - yB x ).
Equations (9-158) have the form (9-153) if
U = | (zyB x + xzBy + yxB,). (9-159)
It is, in fact, possible to express the electromagnetic force in the form
(9-153) for any electric and magnetic field. The electromagnetic force
on a particle of charge q is given by Eq. (3-283):
F = gE + ^ v X B. (9-160)
c
390 lagrange's equations [chap. 9
It is shown in electromagnetic theory* that for any electromagnetic field,
it is possible to define a scalar function 4>{x, y, z, t) and a vector function
A(x, y, z, t) such that
E= _ V 4,_i^, (9-161)
c at
B = V X A. (9-162)
The function <j> is called the scalar potential, and A is called the vector
potential. If these expressions are substituted in Eq. (9-160), we obtain
F = _ gT0 _ 1 d A + 2 v x ( V x A). (9-163)
c at c
The last term can be rewritten using formula (3-35) for the triple cross
product :
F=-gV*-^-^vVA + ? V(v-A). (9-164)
c di c c
[The components of v are (x, y, z) and are independent of x, y, z, so that v
is not differentiated by the operator V.] The two middle terms can be
combined according to Eq. (8-113):
F= -qV4>- g ^ + g V(vA), (9-165)
c at c
where dk/dt is the time derivative of A evaluated at the position of the
moving particle. It may now be verified by direct computation that the
potential function
U = q4> - ^ v-A, (9-166)
when substituted in Eqs. (9-153), with q u q 2 , qz — x, y, z, yields the
components of the force F given by Eq. (9-165). It is also easy to show
that the energy E denned by Eq. (9-156) with L = T — U is
E = T + q<j>. (9-167)
If A and <£ are independent of t, then L is independent of t in a fixed
coordinate system and the energy E is constant, a result derived by more
elementary methods in Section 3-17 [Eq. (3-288)].
When there is a velocity dependent potential, it is customary to define
the momentum in terms of the Lagrangian function, rather than in terms
* See, e.g., Slater and Frank, Electromagnetism. New York: McGraw-Hill
Book Co., 1947. (Page 87.)
Px =
mx -f- - A x ,
c
Py =
m v + f A v>
Pz =
mz + - A z .
c
9-9] lagrange's equations for the vibrating string 391
of the kinetic energy:
If the potential is not velocity dependent, then this definition is equivalent
to Eq. (9-23). In any case, it is dL/dq^ whose time derivative occurs in
the Lagrange equation for q k , and which is constant if q k is ignorable.
In the case of a particle subject to electromagnetic forces, the momentum
components p x , p v , p z will be, by Eqs. (9-168) and (9-166),
(9-169)
The second terms play the role of a potential momentum.
It appears that gravitational forces, electromagnetic forces, and indeed
all the fundamental forces in physics can be expressed in the form (9-153),
for a suitably chosen potential function U. (Frictional forces we do not
regard as fundamental in this sense, because they are ultimately reducible
to electromagnetic forces between atoms, and hence are in principle also
expressible in the form (9-153) if we include all the coordinates of the
atoms and molecules of which a physical system is composed.) Therefore
the equations of motion of any system of particles can always be expressed
in the Lagrangian form (9-155), even when velocity dependent forces are
present. It appears that there is something fundamental about the form
of Eqs. (9-155). One important property of these equations, as we have
already noted, is that they retain the same form if we substitute any new
set of coordinates for q lt . . . , q f . This can be verified by a straightforward,
if somewhat tedious, calculation. Further insight into the fundamental
character of the Lagrange equations must await the study of a more ad-
vanced formulation of mechanics utilizing the calculus of variations, which
is beyond the scope of this book.*
9-9 Lagrange's equations for the vibrating string. The Lagrange
method can be extended also to the motion of continuous media. We shall
consider only the simplest example, the vibrating string. Using the nota-
tion of Section 8-1, we could take u{x) as a set of generalized coordinates
analogous to q k . In place of the subscript k denoting the various degrees
of freedom, we have the position coordinate x denoting the various points
* See, e.g., H. Goldstein, Classical Mechanics. Reading, Mass.: Addison-
Wesley, 1950. (Chapter 2.)
392 lagbange's equations [chap. 9
on the string. The number of degrees of freedom is infinite for an ideal
continuous string. The generalization of the Lagrange method to deal
with a continuous index x denoting the various degrees of freedom intro-
duces mathematical complications which we wish to avoid here.* There-
fore we make use of the possibility of representing the function u(x) as a
Fourier series.
According to the Fourier series theorem quoted in Section 8-2, if the
string is tied at the ends x = 0, I, we can represent its position u(x) by
the series (8-24) :
«(*) = f>sin*P- (9-170)
The coefficients qu are given by Eq. (8-25)
.1
2 /
= 7 / u(x) sin ^ dx, k = 1, 2, 3, . . . . (9-171)
l Jo i
Since the coefficients g* give a complete description of the position of the
string, they represent a suitable set of generalized coordinates. When the
string vibrates, the coordinates gj. become functions of t:
u(x,t) = f>*(«)sin^- (9-172)
We have still an infinite number of coordinates q^, but they depend on the
discrete subscript k and can be treated exactly like the generalized coordi-
nates considered earlier in this chapter. Since the string could in principle
be treated as a system with a very large number of particles, and since we
are allowed to describe the system by any suitable set of generalized co-
ordinates, we need only express the Lagrangian function in terms of the
coordinates qk in order to write down the equations of motion.
We first need to calculate the kinetic energy, which is evidently
r = /o**(l) V (9 " 173)
If we differentiate Eq. (9-172) with respect to t and square, we obtain
©^ggM^in^sin^. (9-174)
* For a treatment of this problem, see H. Goldstein, op. cit. (Chapter 11.)
9-9] lagrange's equations for the vibrating string 393
We now multiply by %<r dx and integrate from to I term by term.* Since
/.
'0
the result obtained is
farx . jirx , \U, j = k,
i —j— sin J -r r - dx = \
0, j * k,
sin — j— sin —f- ax = {- - ' / Q 17 r-\
I I In i ■ ^ h (9-175)
T = J2 && (9-176)
4=1
We next calculate the generalized force Q k . If coordinate q k increases by
Sq k , while the rest are held fixed, a point x on the string moves up a dis-
tance given by Eq. (9-170) :
8u = Sq k sin ^ • (9-177)
The upward force on an element dx of string is given by Eq. (8-3). The
work done is therefore
Jo dx\ dx/
w = QkS «» = J irxVix)* udx - c 9 - 178 )
We substitute for du/dx from Eq. (9-170), and for Su from Eq. (9-177),
and integrate term by term, to obtain (assuming t constant) :
Qk = -ih (jrf qk. (9-179)
The forces Q k are obviously derivable from the potential energy function,
V=±ilr(^f) 2 ql (9-180)
It will be instructive to calculate V directly by calculating the work done
against the tension t in moving the string from its equilibrium position to
the position u(x). At the same time we shall verify that this work is in-
dependent of how we move the string to the position u(x). Let u(x, t) be
the position of the string at any time t while the string is being moved to
* In order to differentiate and integrate infinite series term by term, and to
rearrange orders of summation, as we shall do freely in this section, we must
require that the series all converge uniformly. This will be the case if u{x, t)
and its derivatives are continuous functions. (For a precise statement and
derivation of the conditions for manipulation of infinite series, see a text on
advanced calculus, e.g., W. Kaplan, Advanced Calculus. Reading, Mass.:
Addison- Wesley, 1952. Chapter 6.)
394 lagkange's equations [chap. 9
u(x). [The function u(x, t) is not necessarily a solution of the equation of
motion, since we wish to consider an arbitrary manner of moving the string
from u = to u = u(x).] At t = 0, the string is in its equilibrium
position:
u(x, 0) = 0. (9-181)
Let t = ti be the time the string arrives at its final position:
u(x, tj) = u(x). (9-182)
The work done against the vertical components of tension [Eq. (8-3)]
during the interval dt is
f d_ ( du\ (du \ dx
~~ Jx=o dx \ dx/ \dt )
We integrate by parts, remembering that u and du/dt are at x — 0, I:
fl 2
,„ / du du , ,.
dV = I t— -rr^- dx dt
J x =o dxotax
The total work done is then
dV
where in the last expression, u = u{x) corresponds to the final position of
the string. The result depends only on the final position of the string — an
independent proof that the tension forces are conservative.
The work done against the tension is stored as potential energy in the
stretched string. By substituting in Eq. (9-184) from Eq. (9-170), we
again can obtain Eq. (9-180). In Eq. (8-61) for a string of particles, the
right member contains two terms that represent the vertical components
of force between adjacent pairs of particles. A third way of deriving the
potential energy is to find the potential energy function between a pair of
9-9] lageange's equations for the vibrating string 395
particles which yields this force. It must then be shown that, when this
is summed over all pairs of adjacent particles, the result approaches Eq.
(9-184) in the limit h -> 0.
The Lagrangian function for the vibrating string can now be written as
L = T - V = £ [fat* ~ & (t) 2 «*] • ( 9 - 185 )
The resulting Lagrange equation for q k is
ilaqk + ¥t (jr) qk = 0, (9-186)
whose general solution is
qk = A k cos u k t + B k sin u k t, (9-187)
-tG)""-*
This result can be substituted in Eq. (9-172) to obtain the solution
u(x, t) — 2_j \A k sin —j- cos u k t + B k sin -y- sin u k t) > (9-189)
which is in agreement with Eq. (8-23). If u = u {x) and du/dt = v (x)
are given at t = 0, we can use Eqs. (9-171) and (9-187) to find the
constants A k , B k :
where
2 / kirx
A k = q k (0) = j u o(x) sin— —dx,
I
D q k (0) 2 / . . . kwx ,
B k = 2-^— = — : / v {x) sm—r-dx,
Wj; U) k l JO I
(9-190)
in agreement with Eqs. (8-25).
The coordinates q k defined by Eqs. (9-170) and (9-171) are called the
normal coordinates for the vibrating string. Each coordinate evidently
represents one normal mode of vibration. The normal coordinates are also
very useful in treating the case where a force f(x, I) is applied along the
string (see Problem 26 at the end of this chapter). Mathematically, the
normal coordinates have the property that the Lagrangian L becomes a
sum of terms, each term involving only one degree of freedom. Thus
in normal coordinates the problem is subdivided into separate problems,
one for each degree of freedom.
396 lagrange's equations [chap. 9
It was, of course, rather fortunate that the coordinates g* which were
chosen at the beginning of the problem turned out to be the normal co-
ordinates. In general, this does not happen. For example, consider a
string whose density varies along its length according to
o - = o"o + asin-p- (9-191)
This string is heaviest near its center. We will use the same coordinates
q k as defined by Eqs. (9-170) and (9-171). We substitute Eqs. (9-191)
and (9-172) in Eq. (9-173) and, instead of Eq. (9-176), we obtain (after
some calculation),
T = E E %T kj q k q h (9-192)
&=i j— i
where
Ala k 2
T ki = &(To + — 4fc2 _ x ' if * = h
«V» - - £ [ < M . j) ,- 1 &- a ,-n ' i!k " * »""»
and k, j are both even or both odd; otherwise
T ki = 0.
For this string, the q k 's are evidently not normal coordinates. In the
Lagrange equations the q k 's, with k even, are all coupled together, as are
those with k odd. The problem is then much more difficult, and a solution
will not be attempted here.
9-10 Hamilton's equations. The discussion in this section will be
restricted to mechanical systems obeying Lagrange's equations in the form
(9-57). The Lagrangian L is a function of the coordinates qk, of the veloci-
ties q k , and perhaps of t. The state of the mechanical system at any time,
that is, the positions and velocities of all its parts, is specified by giving the
generalized coordinates and velocities qk, qk- Lagrange's equations are
second-order equations which relate the accelerations q k to the coordinates
and velocities. The state of the system could equally well be specified by
giving the coordinates q k and the momenta p k defined by Eq. (9-168) :
Pk = Wk' fc= 1,2> ""' / ' (9_194)
These equations specify pk in terms of qi, . . . , g/; fa, . . . , fa. They can,
in principle, be solved for q k in terms of qi, . . . , q/; p u . . . , p/.
9-10] Hamilton's equations 397
It is an interesting exercise to try to write equations of motion in terms
of the coordinates qk and momenta pk- Note first that, by use of the
definition (9-194) and the equations of motion (9-57), we have
= 22 (Pk dq k + Pk dq k ) + -rr dt. (9-195)
We next define a function H(q lt . . . , gy; p lf . . . , p f ; t) by
/
H = J2 Vkik - L, (9-196)
where for the velocities gj we substitute their expressions in terms of co-
ordinates and momenta. Then we have
dH = J2 (& dp k - p k dq k ) - ~ dt. (9-197)
The definition (9-196) is chosen so that dH depends explicitly upon dp k ,
dqk, and dt. By inspection of Eq. (9-197), we see that
tiff riff
and
dH = _dL
dt dt '
(9-199)
Equations (9-198) are the desired equations of motion that express q k and
pk in terms of the coordinates and momenta.
Equations (9-198) are Hamilton's equations of motion for a mechanical
system. The function H, defined by Eq. (9-196), is called the Hamiltonicm
function. We see from Eq. (9-124) that when V is a function only of the
coordinates, for a stationary coordinate system, H is just the total energy
expressed in terms of coordinates and momenta. For a moving coordinate
system, where T is given by Eq. (9-13), the Hamiltonian is
H = T 2 + V - T , (9-200)
with T 2 expressed in terms of coordinates and momenta. According to
Section 9-8, H will also be the total energy in a stationary coordinate
system when electromagnetic forces are present.
398 lagrange's equations [chap. 9
When L does not contain the time explicitly, neither does H according
to Eq. (9-199), as is also obvious from the way in which H was denned.
According to Eq. (9-125), H is a constant of the motion in this case. This
can also be proved directly from Eqs. (9-198), since it is easy to show that
^ = ^ , (9-201)
dt dt
as the reader may verify.
If any coordinate qk does not appear explicitly in H, then Eqs. (9-198)
give
p k = a constant, (9-202)
in agreement with Eq. (9-118). Since H does not contain q k , we may take
p k as a given constant, and the 2(/ — 1) equations (9-198) for the other
coordinates and momenta are then the Hamiltonian equations for a sys-
tem of / — 1 degrees of freedom. Thus degrees of freedom corresponding
to coordinates that do not appear in H simply drop out of the problem.
This is the origin of the term "ignorable coordinate. " After the remaining
equations of motion have been solved for the nonignorable coordinates and
momenta, any ignorable coordinate is given by Eqs. (9-198) as an integral
over t:
q k (t) = «*«>) + / ff dt. (9-203)
JO °Pk
Hamilton's equations are simply a new formulation of Newton's laws
of motion. In simple cases, they reduce to equations which could have been
written immediately from Newton's laws. In the harmonic oscillator, for
example, with coordinate x, the momentum is
p = mi. (9-204)
The Hamiltonian function is therefore
H = T + V = ^ + ikx 2 . (9-205)
Equations (9-198) become
x = 2-, p = —fee. (9-206)
m
The first of these is the definition of p, and the second is Newton's equation
of motion.
Although they are of comparatively little value as a means of writing
the equations of motion of a system, Hamilton's equations are important
for two general reasons. First, they provide a useful starting point in
setting up the laws of statistical mechanics and of quantum mechanics.
9-11] liouville's theorem 399
Hamilton orginally developed his equations by analogy with a similar
mathematical formulation which he had found useful in optics. It is not
surprising that Hamilton's equations should form the starting point for
wave mechanics! Second, there are a number of methods of solution of
mechanical problems based on Hamilton's formulation of the equations of
motion. It is clear from the way in which they were derived that Hamil-
ton's equations (9-198), like Lagrange's equations, are valid for any set
of generalized coordinates qi, . . . , q/ together with the corresponding
momenta Pi, . . . , p/, defined by Eq. (9-194). In fact Hamilton's equations
are valid for a much wider class of coordinate systems obtained by de-
fining new coordinates and momenta as certain functions of the original
coordinates and momenta. This is the basis for the utility of Hamilton's
equations in the solution of mechanical problems. A further discussion of
these topics is beyond the scope of this book.* We will, however, prove
one general theorem in the next section which gives some insight into the
importance of the variables pk and qu.
9-11 Liouville's theorem. We may regard the coordinates q u . . . , qy
as the coordinates of a point in an /-dimensional space, the configuration
space of the mechanical system. To each point in the configuration space
there corresponds a configuration of the parts of the mechanical system.
As the system moves, the point Qi, . . . , Q/ traces a path in the configura-
tion space. This path represents the history of the system.
If we wish to specify both the configuration and the motion of a system
at any given instant, we must specify the coordinates and velocities, or
equivalently, the coordinates and momenta. The 2/-dimensional space
whose points are specified by the coordinates and momenta Qi, . . • , Q/',
Pi, . . . , pf is called the phase space of the mechanical system. As the sys- .
tem moves, the phase point gi, ...,<?/; Pi, •••, p/ traces out a path in the
phase space. The velocity of the phase point is given by Hamilton's
equations (9-198).
Each phase point represents a possible state of the mechanical system.
Let us imagine that each phase point is occupied by a "particle" which
moves according to the equations of motion (9-198). These particles
trace out paths that represent all possible histories of the mechanical
system. The theorem of Liouville states that the phase "particles" move
as an incompressible fluid. More precisely, the phase volume occupied by
a set of "particles" is constant.
To prove Liouville's theorem, we make use of theorem (8-121) gener-
alized to a space of 2/ dimensions. We may either generalize the argument
which led to Eq. (8-116), or we may use the generalization of Gauss'
* See H. Goldstein, op. cit. (Chapters 7, 8, 9.)
4Q0 lagrange's equations [chap. 9
divergence theorem which is valid in any number of dimensions. In either
case, we have for a volume V in phase space, moving with the "particles":
dV
dt
- /•; •/ S (t +*£)*■- * *• • ■ • *" fr*"
which is Eq. (8-121) written for the 2/-dimensional phase space. We now
substitute the velocities from Hamilton's equations (9-198) :
(9-208)
This is Liouville's theorem, and it should be noted that this theorem holds
even when H depends explicitly on t.
In the case of a harmonic oscillator, the phase space is a plane with co-
ordinate axes x and p. The phase points move around ellipses H = con-
stant, given by Eq. (9-205), and with velocities as given by Eq. (9-206).
According to Liouville's theorem, the motion is that of a two-dimensional
incompressible fluid. In particular, a set of points that lie in a region of
area A will at any later time lie in another region of area A.
Liouville's theorem makes the coordinates and momenta more useful
for many purposes than coordinates and velocities. Because of this
theorem, the concept of phase space is an important tool in statistical
mechanics. Imagine a large number of mechanical systems identical to a
given one, but with different initial conditions. Let each system be repre-
sented by a point in their common phase space, and let these points move
according to Hamilton's equations. The statistical properties of this col-
lection of systems may be specified at any time t by giving the density
P(<Zi. • • • > If) Pi, ■ • • > Pf', m the phase space of system points per unit
volume. Liouville's theorem implies that the density p in the immediate
neighborhood of any system point must remain constant as that point
moves through the phase space. (Why?) If we define statistical equi-
librium as a distribution in which p is constant in time at each fixed point
in the phase space, then clearly the necessary and sufficient condition for
equilibrium is that p be uniform along the flow lines of the system points.
(Why?)
We have been able in this section to give only a narrow glimpse of the
power of the Hamiltonian methods.
401
Problems
1. Coordinates u, w are defined in terms of plane polar coordinates r, by the
equations
u = In (r/a) — cot f ,
w = In (r/a) + 6 tan J",
where a and f are constants. Sketch the curves of constant u and of constant w.
Find the kinetic energy for a particle of mass m in terms of u, w, u, ib. Find
expressions for Q u , Q m in terms of the polar force components F r , Ft. Find p u , p w .
Find the forces Q a , Q w required to make the particle move with constant speed s
along a spiral of constant u = uo.
2. Two masses' mi and m.2 move under their mutual gravitational attraction
in a uniform external gravitational field whose acceleration is g. Choose as
coordinates the cartesian coordinates X, Y, Z of the center of mass (taking Z in
the direction of g), the distance r between mi and wi2, and the polar angles
and <p which specify the direction of the line from mi to wi2- Write expres-
sions for the kinetic energy, the six forces Qx, . . . ,Q V , and the six momenta.
Write out the six Lagrange equations of motion.
3. (a) Set up the expression for the kinetic energy of a particle of mass m in
terms of plane parabolic coordinates /, h, as defined in Problem 13 of Chapter 3.
Find the momenta p/ and p*. (b) Write out the Lagrange equations in these
coordinates if the particle is not acted. on by any force. .
4. (a) Find the forces Q; and Qh required to make the particle in Problem 3
move along a parabola / = /o = a constant, with constant generalized velocity
ft, = ho, starting from h = at t = 0. (b) Find the corresponding forces F x
and F v relative to a cartesian coordinate system.
5. (a) Set up the Lagrange equations of motion in spherical coordinates r, 6, <p,
for a particle of mass m subject to a force whose spherical components are
F r , Fe, F v .
(b) Set up Lagrange equations of motion for the same particle in a system of
spherical coordinates rotating with angular velocity co about the 3-axis.
(c) Identify the generalized centrifugal and coriolis forces 'Q r ', 'Qe', and 'Q v '
by means of which the equations in the rotating system can be made to take the
same form as in the fixed system. Calculate the spherical components 'F r ', 'Ff',
'F v ' of these centrifugal and coriolis forces, and show that your results agree
with the expressions derived in Chapter 7.
6. Set up the Lagrangian function for the mechanical system shown in
Fig. 4-16, using the coordinates x, x\, X2 as shown. Derive the equations of
motion, and show that they are equivalent to the equations that would be
written down directly from Newton's law of motion.
7. Choose suitable coordinates and write down the Lagrangian function for
the restricted three-body problem. Show that it leads to the equations of mo-
tion obtained in Section 7-6.
8. Masses m and 2m are suspended from a string of length h which passes
over a pulley. Masses 3m and 4m are similarly suspended by a string of length
402 lagrange's equations [chap. 9
h over another pulley. These two pulleys hang from the ends of a string of length
h over a third fixed pulley. Set up Lagrange's equations, and find the accelera-
tions and the tensions in the strings.
9. A massless tube is hinged at one end. A uniform rod of mass m, length I,
slides freely in it. The axis about which the tube rotates is horizontal, so that the
motion is confined to a plane. Choose a suitable set of generalized coordinates,
one for each degree of freedom, and set up Lagrange's equations.
10. Set up Lagrange's equations for a uniform door whose axis is slightly out
of plumb. What is the period of small vibrations?
11. A double pendulum is formed by suspending a mass mi by a string of
length h from a mass m\ which in turn is suspended from a fixed support by a
string of length l\. (a) Choose a suitable set of coordinates, and write the
Lagrangian function, assuming the double pendulum swings in a single vertical
plane.
(b) Write out Lagrange's equations, and show that they reduce to the equa-
tions for a pair of coupled oscillators if the strings remain nearly vertical.
(c) Find the normal frequencies for small vibrations of the double pendulum.
Describe the nature of the corresponding vibrations. Find the limiting values
of these frequencies when wu >J> W2, and when W2 » mi. Show that these
limiting values are to be expected on physical grounds by considering the nature
of the normal modes of vibration when either mass becomes vahishingly small.
12. A ladder rests against a smooth wall and slides without friction on wall
and floor. Set up the equation of motion, assuming that the ladder maintains
contact with the wall. If initially the ladder is at rest at an angle a with the
floor, at what angle, if any, will it leave the wall?
13. One end of a uniform rod of mass M makes contact with a smooth vertical
wall, the other with a smooth horizontal floor. A bead of mass m and negligible
dimensions slides on the rod. Choose a suitable set of coordinates, set up the
Lagrangian function, and write out the Lagrange equations. The rod moves in a
single vertical plane perpendicular to the wall.
14. A ring of mass M rests on a smooth horizontal surface and is pinned at a
point on its circumference so that it is free to swing about a vertical axis. A
bug of mass m crawls around the ring with constant speed, (a) Set up the equa-
tions of motion, taking this as a system with two degrees of freedom, with the
force exerted by the bug against the ring to be determined from the condition
that he moves with constant speed.
(b) Now set up the equation of motion, taking this as a system with one de-
gree of freedom, the bug being constrained to be at a certain point on the ring
at each instant of time. Show that the two formulations of the problem are
equivalent.
15. A pendulum bob of mass m is suspended by a string of length I from a
point of support. The point of support moves to and fro along a horizontal z-axis
according to the equation
x = a cos ciit.
Assume that the pendulum swings only in a vertical plane containing the z-axis.
Let the position of the pendulum be described by the angle which the string
PROBLEMS
403
makes with a line vertically downward, (a) Set up the Lagrangian function and
write out the Lagrange equation.
(b) Show that for small values of 6, the equation reduces to that of a forced
harmonic oscillator, and find the corresponding steady-state motion. How does
the amplitude of the steady-state oscillation depend on m, I, a, and co?
16. A pendulum bob of mass m is suspended by a string of length I from a car
of mass M which moves without friction along a horizontal overhead rail. The
pendulum swings in a vertical plane containing the rail, (a) Set up the Lagrange
equations, (b) Show that there is an ignorable coordinate, eliminate it, and dis-
cuss the nature of the motion by the energy method.
17. Find the tension in the string for the spherical pendulum discussed in
Section 9-7, as a function of E, p ? , and 8. Determine, for a given E and p v ,
the angle Si at which the string will collapse.
18. A particle of mass m slides over the inner surface of an inverted cone of
half-angle a. The apex of the cone is at the origin, and the axis of the cone ex-
tends vertically upward. The only force acting on the particle, other than the
force of constraint, is the force pf gravity, (a) Set up the equations of motion,
using as coordinates the horizontal distance p of the particle from the axis, and
the angle <p measured in a horizontal circle around the cone. Show that <p is
ignorable, and discuss the motion by the method of the effective potential.
(b) For a given radius po, find the angular velocity <f>o of revolution in a hori-
zontal circle, and the angular frequency o> of small oscillations about this cir-
cular motion. Show that the small oscillations are a wobbling or an up-and-
down spiraling motion, depending on whether the angle a is greater than or less
than the angle
-i J_
sin
19. A flyball governor for a steam engine is shown in Fig. 9-11. Two balls,
each of mass m, are attached by means of four hinged arms, each of length I,
to sleeves which slide on a vertical rod. The upper sleeve is fastened to the rod;
the lower sleeve has mass M and is free to slide up and down the rod as the balls
Fig. 9-11. A flyball governor.
404 laghange's equations [chap. 9
move out from or toward the rod. The rod-and-ball system rotates with con-
stant angular velocity w. (a) Set up the equation of motion, neglecting the weight
of the arms and rod. Discuss the motion by the energy method.
(b) Determine the value of the height z of the lower sleeve above its lowest
point as a function of o> for steady rotation of the balls, and find the frequency
of small oscillations of z about this steady value.
20. Discuss the motion of the governor described in Problem 19 if the shaft
is not constrained to rotate at angular velocity «, but is free to rotate, without
any externally applied torque, (a) Find the angular velocity of steady rotation
for a given height z of the sleeve, (b) Find the frequency of small vibrations
about this steady motion, (c) How does this motion differ from that of Prob-
lem 19?
21. A rectangular coordinate system with axes x, y, z is rotating with uniform
angular velocity w about the z-axis. A particle of mass m moves under the action
of a potential energy V(x, y, z). (a) Set up the Lagrange equations of motion.
(b) Show that these equations can be regarded as the equations of motion of a
particle in a fixed coordinate system acted on by the force — VF, and by a force
derivable from a velocity dependent potential U. Hence find a velocity de-
pendent potential for the centrifugal and coriolis forces. Express U in spherical
coordinates r, 6, <p, f, 6, <p, and verify that it gives rise to the forces 'Q r ', 'Qt, 'Q v '
found in Problem 5.
22. Show that a uniform magnetic field B in the z-direction can be represented
in cylindrical coordinates (Fig. 3-22) by the vector potential
A = \Bp m.
Write out the Lagrangian function for a particle in such a field. Write down
the equations of motion, and show that there are three constants of the motion.
Compare with Problem 49 of Chapter 3.
23. The kinetic part of the Lagrangian function for a particle of mass m in
relativistic mechanics is
L k mc 2 [l - (y/c) 2 ] 1/2 .
Show that this gives the proper formula (4-75) for the components of momentum.
Show that if the potential function for electromagnetic forces Eq. (9-166) is
subtracted, and if A and <f> do not depend explicitly on t, then T -j- q<f> is con-
stant, with T given by formula (4-74).
*24. Show by direct calculation that if Eqs. (9-155) hold for some function
L(qi, . . . , <?/; 51, ... , fa; t), and we introduce new coordinates q*, . . . , q*,
where
qk = Mqi, ■ ■ ■ ,qf,t), k = 1, . . . , /,
then
lit w* a * — ' ' — *> • • ■ >/>
at dq t dq t
where L*(q%, . . . , q*; q*, . . . , q*; t) = L{qi, ...,qf, qi, ... , q f ; t) is obtained
by substitution of /*(?*, . . . , q*; t) for q k .
PROBLEMS 405
25. Derive formula (9-184) by writing down a potential energy which gives
the interparticle forces for the string of particles studied in Section 8-4, and
passing to the limit h —* 0.
26. A stretched string is subject to an externally applied force of linear density
f(x, t). Introduce normal coordinates qk, and find an expression for the gen-
eralized applied force Qk(t). Use the Lagrangian method to solve Problem 6(a),
Chapter 8.
27. Solve Problem 7, Chapter 8, by using the coordinates g* defined by
Eqs. (9-170) and (9-171).
28. Write down the Hamiltonian function for the spherical pendulum. Write
the Hamiltonian equations of motion, and derive from them Eq. (9-136).
*29. Work out the relativistic Hamiltonian function for a particle subject to
electromagnetic forces, using the Lagrangian function given in Problem 23.
Write out the Hamiltonian equations of motion and show that they are equiva-
lent to the Lagrange equations.
30. Work out the Hamiltonian function H(qk, pic) for the vibrating string,
starting from Eq. (9-185). Write down the equations which relate the momenta
Pk to the function u(x, t) which describes the motion of the string. Hence show
that H = T + V, with T and V given by Eqs. (9-173) and (9-184).
31. Write down the Hamiltonian function for Problem 2. Write out Hamil-
ton's equations. Identify the ignorable coordinates and show that there remain
two separate one degree of freedom problems, each of which can be solved (in
principle) by the energy method. What are the corresponding two potential-
energy functions?
32. A beam of electrons is directed along the z-axis. The electrons are uni-
formly distributed over the beam cross section, which is a circle of radius oo,
and their transverse momentum components (p x , Pv) are distributed uniformly
in a circle (in momentum space) of radius po- If the electrons are focused by
some lens system so as to form a spot of radius oi, find the momentum distribu-
tion of electrons arriving at the spot.
33. A group of particles all of the same mass m, having initial heights and
vertical momenta lying in the square — a < z < a, — b < p < b, fall freely
in the earth's gravitational field for a time t. Find the region in the phase space
within which they lie at time t, and show by direct calculation that its area is still
4o6.
34. In an electron microscope, electrons scattered from ah object of height
zo are focused by a lens at distance Do from the object and form an image of
height zi at a distance Dj behind the lens. The aperture of the lens is A. Show
by direct calculation that the phase area in the (z, p z ) phase plane occupied by
electrons leaving the object (and destined to pass through the lens) is the same
as the phase area occupied by electrons arriving at the image. Assume that
zo <3C Do and zi <C Dr.
CHAPTER 10
TENSOR ALGEBRA. INERTIA AND STRESS TENSORS
In this chapter we shall develop the algebra of linear vector functions,
or tensors, as a mathematical tool which is useful in treating many prob-
lems. In particular, we shall need tensors in the study of the general
motion of a rigid body and in the formulation of the concept of stress in
a solid, or in a viscous fluid.
10-1 Angular momentum of a rigid body. The equation of motion for
the rotation of a rigid body is given by Eq. (5-5) and restated here:
ft = N ' < 1(H >
where L is the angular momentum and N is the torque about a point P
which may be either fixed or the center of mass of the body. In Section
5-2 we studied the rotation of a rigid body about a fixed axis. In order
to treat the general problem of the rotation of a body about a point P,
we must find the relation between the angular momentum vector L and
the angular velocity vector u.
Consider a body made up of point masses m k situated at points r k
relative to an origin of coordinates at P. We have shown in Section 7-2
that the most general motion of the body about the point P is a rotation
with angular velocity u, and that the velocity Vfc of each particle in the
body is given by
v fc = to X r k . (10-2)
We sum the angular momentum given by Eq. (3-142) over all particles:
L = "22 in k r k X v&
k=l
N
= X) m * r * X (« X r *)- (10-3)
Equation (10-3) expresses L as a function of a>, L(«). By substitution in
Eq. (10-3) it is readily verified that the function L(«), for any two vectors
w, w', and any scalar c, satisfies the following relations:
L(cw) = cL(«), (10-4)
L(» + «') = L(») + L(<o'). (10-5)
406
10-2] TENSOR ALGEBRA 407
A vector function L(w) with the properties (10-4), (10-5) is called a
linear vector function. Linear vector functions are important because they
occur frequently in physics, and because they have simple mathematical
properties.
In order to develop an analogy between Eq. (10-3) and Eq. (5-9) for
the case of rotation about an axis, we make use of Eq. (3-35) :
N
I
4=1
L = ^2 [m k rlw — m k i k {r k ■ «)]. (10-6)
The factor u is independent of k and can be factored from the sum over
the first term. In a purely formal way, we may also factor « from the sum
over the second term:
L = (X) m * r *) w — (S m * r fc r fc) • «• (10-7)
The second term has no meaning, of course, since the juxtaposition Tkik
of two vectors has not yet been denned. We shall try to supply a mean-
ing in the next section.
10-2 Tensor algebra. The dyad product AB of two vectors is denned
by the following equation, where C is any vector:
(AB) • C = A(B • C). (10-8)
The right member of this equation is expressed in terms of products de-
fined in Section 3-1. The left member is, by definition, the vector given by
the right member. Note that the dyad AB is defined only in terms of its dot
product with an arbitrary vector C. We can readily show, from definition
(10-8), that multiplication of a vector by a dyad is a linear operation in
the sense that
(AB) • (cC) = c[(AB) • C], (10-9)
(AB) • (C + D) = (AB) • C + (AB) • D. (10-10)
For fixed vectors A, B, the dyad AB therefore defines a linear vector func-
tion F(C) :
F(C) = (AB) • C. (10-11)
The dyad AB is an example of a linear vector operator, that is, it represents
an operation which may be performed on any vector C to yield a new
vector (AB) • C, which is a linear function of C.
408 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
A linear vector operator is also called a tensor * Tensors will be repre-
sented by sans-serif boldface capitals, A, B, C, etc. We may for example
let T be the tensor represented by the dyad AB:
T = AB. (10-12)
The meaning of the tensor T is specified by the definition,!
T • C = A(B • C), (10-13)
which gives the result of applying T to any vector C. We can form more
general linear vector operations by taking sums of dyads. The sum of
two dyads, or tensors S, T, is denned as follows:
(S + T) • C = S • C + T • C. (10-14)
Note that all definitions of algebraic operations on tensors, like the above
definition of (S + T), are formulated in terms of the application of the
tensors to an arbitrary vector C. The sum of one or more dyads is called
a dyadic. According to the definition (10-14), the dyadic (AB + DE)
operating on C yields the vector
(AB + DE) • C = A(B • C) + D(E • C). (10-15)
We can readily show that the sum of two linear operators is a linear
operator; therefore dyadics are also linear vector operators and we have for
any dyadic or tensor T,
T • (cC) = c(T • C), (10-16)
T • (C + D) = T • C + T • D. (10-17)
The linearity relations (10-16), (10-17), together with the definition
(10-14), guarantee that dyad products, sums of tensors, and dot products
of tensors with vectors satisfy all the usual algebraic rules for sums and
products. We can also define a dot product of a dyad with a vector on the
left in the obvious way,
C • (AB) = (C • A)B, (10-18)
* More precisely, a linear vector operator may be called a second-rank tensor,
to distinguish it from third- and higher-rank tensors obtained as linear combina-
tions of triads ABC, etc. We shall be concerned in this book only with second-
rank tensors, which we shall refer to simply as tensors.
t The result of applying a tensor T to a vector C is often denoted by TC,
without the dot. We shall use the dot throughout this book.
10-2] TENSOR ALGEBRA 409
and correspondingly for sums of dyads. Note that the dot product of a
dyadic with a vector is not commutative;
T • C = C • T (10-19)
does not hold in general. We can define, in an obvious way, a product cT
of a tensor by a scalar, with the expected algebraic properties (see Problem
1).
A very simple tensor is given by the dyadic
1 = ii + j j + (10-20)
where i, j, k are a set of perpendicular unit vectors along x-, y-, and z-
axes. We calculate, using the definitions (10-14) and (10-8),
1 • A = \A X + ]A V + kA z = A. (10-21)
The tensor 1 is called the unit tensor; it may be defined as the operator
which, acting on any vector, yields that vector itself. Evidently 1 is one
of the special cases for which
1A = A1. (10-22)
If c is any scalar, the product cl is called a constant tensor, and has the
property
(cl) • A = A • (cl) = cA. (10-23)
Using the definitions above, we can now write Eq. (10-7) in the form
L = 1 • to, (10-24)
where I is the inertia tensor of the rigid body, defined by
N
I
4=1
I = g {m k rl\ - m k r k T k ). (10-25)
The inertia tensor I is the analog, for general rotations, of the moment of
inertia for rotations about an axis. Note that L and a are not in general
parallel. We will study the inertia tensor in more detail after we have
developed the necessary properties of tensors.
If we write all vectors in terms of their components,
C = C x i + C v j + CM, (10-26)
then it is clear that by multiplying out dyad products and collecting terms,
any dyadic can be written in the form:
410 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
T = T xx ii + T xy ij + T xz ik
+ T vx ji + Tyyjj + T ye jb
+ TJU + T zy kj + T zz kk. (10-27)
Just as any vector A can be represented by its three components (A x , A y ,
A z ), so any dyadic can be specified by giving its nine components T xx ,
. . . , T zz . These may conveniently be written in the form of a square array
or matrix:
/ -* XX ■* X
(10-28)
As an example, the reader may verify that the components of the inertia
tensor (10-25) are
N N
Ixx = X) m k (yl + zl), I xv = — ^ m k x k y k , etc. (10-29)
fc=i fc=i
In order to simplify writing the tensor components, it often will be con-
venient to number the coordinate axes xi, x 2 , £3 instead of using x, y, z:
x = x h y — x 2 , z = x 3 . (10-30)
We shall write the corresponding unit vectors as e*:
i = e 1( j = e 8 , k = e 3 . (10-31)
Equations (10-26) and (10-27) can now be written as
C = X) Ctfii, (10-32)
i—X
and
3
T = J^ Tifi^i- (10-33)
Another advantage of this notation is that it allows the discussion to be
generalized to vectors and tensors in a space of any number of dimensions
simply by changing the summation limit.
By using the definitions of dyad products and sums, we can express the
components of the vector T • C in terms of the components of T and C:
(T-C),-= YiTifij, (10-34)
3=1
10-2] TENSOR ALGEBRA
as the reader should verify. Similarly,
(C • Vi = E OF*
3=1
We note that, by Eq. (10-33),
T ti = e< • (T • e,-) = (e,- • T) • e y .
411
(10-35)
(10-36)
We may omit the parentheses, since the order in which the multiplications
are carried out does not matter.
We can now show that any linear vector function can be represented by
a dyadic. Let F(C) be any linear function of C. Consider first the case
when C is a unit vector ey, and let T t j be the components of F in that case:
(10-37)
F(e,) =
3
Now
any vector C can be written
as
C =
= E c &-
3=1
By use of the linear property of F(C), we have therefore
F(C) =
E F ^' e ^
3-1
=
E c >x&)
3=1
=
3
(10-38)
(10-39)
i, j— i
Thus the components of F(C) can be expressed in terms of the numbers
If we define the dyadic
[F(C)] t - = E Ttfii
3=1
we see from Eqs. (10-40) and (10-34) that
F(C) = T • C.
(10-40)
(10-41)
(10-42)
412 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
Thus the concepts of dyadic and linear vector operator or tensor are
identical, and are equivalent to the concept of linear vector function
in the sense that every linear vector function defines a certain tensor or
dyadic, and conversely.
We can define a dot product of two tensors as follows:
(T • S) • C = T • (S • C). (10^3)
Application of the operator T • S to any vector means first applying S,
and then T. We now calculate in terms of components, using the definition
(10-43),
(T • S) • C = T ■ X) SrftiBj
3 3
i=i j,k=i
= E IE (E T i&>) c*l •*• ( 10 ^ 4 )
i=l L ft=1 \ ]=1 / J
Comparing this result with Eq. (10-34), we see that
3
<J • S)* = IE TijSju. (HM5)
Equation (10-45) also results if we simply evaluate T • S in a formal way
by writing the dot and dyad products and collecting terms:
3
T • S = ^ TijSkie&j • e k ei
ijhl=l
3
= X) TijSjiPfii, (10-46)
ijl—l
and this shows that our definition (10-43) is consistent with the ordinary
rules of algebra. If T, S are written as matrices according to Eq. (10-28),
then Eq. (10-45) is the usual mathematical rule for multiplying matrices.
We can similarly show that the definition (10-14) implies that tensors
are added by adding their component matrices according to the rule:
(T + S) i3 - = Tn + Sij. (10-47)
Sums and products of tensors obey all the usual rules of algebra except
that dot multiplication, in general, is not commutative:
10-2] TENSOR ALGEBRA 413
T + S = S + T, (10-48)
T • (S + P) = T • S + T • P, (10-49)
T • (S • P) = (T • S) • P, (10-50)
1 . T = T • 1 = T, (10-51)
and so on, but
T • S 5* S • T, in general. (10-52)
It is useful to define the transpose V of a tensor T as follows:
T' • C = C ■ T. (10-53)
In terms of components,
Tlj = T it . (10-54)
The transpose is often written T, but the notation T' is preferable for
typographical reasons. The following properties are easily proved:
(T + S)« = T* + $', (10-55)
(T • S) f = S* • T', (10-56)
(¥*)' = T. (10-57)
A tensor is said to be symmetric if
T 1 = T. (10-58)
For example, the inertia tensor, given by Eq. (10-25) is symmetric. For
a symmetric tensor,
Ta = T tJ . (10-59)
A symmetric tensor may be specified by six components; the remaining
three are then determined by Eq. (10-59).
A tensor is said to be antisymmetric if
V = -T. (10-60)
The components of an antisymmetric tensor satisfy the equation
Tji = -Tij. (10-61)
Evidently the three diagonal components Tu are all zero, and if three off-
diagonal components are given, the three remaining components are given
414 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
by Eq. (10-61). An antisymmetric tensor has only three independent
components (in three-dimensional space). An example is the linear operator
defined by
T • C = to X C, (10-62)
where to is a fixed vector. Comparing Eq. (10-62) with Eq. (7-20), we
see that the operator T can be interpreted as giving the velocity of any
vector C rotating with an angular velocity a. Comparing Eq. (10-62)
with Eq. (10-34), we see that the components of T are:
Tn = T22 = T33 = 0,
T21 = — T12 = W3,
(10-63)
T S2 = — T23 = «i,
^13 = —T31 = CO2. t
Since an antisymmetric tensor, like a vector, has three independent com-
ponents, we may associate with every antisymmetric tensor T a vector
to (in three-dimensional space only!) whose components are related to
those of T by Eq. (10-63). The operation T • will then be equivalent to
u X , according to Eq. (10-62).
Given any tensor T, we can define a symmetric and an antisymmetric
tensor by
T. = i(T + TO, (10-64)
T« = i(T - T'), (10-65)
such that
T = T 8 + T„. (10-66)
We saw, in the preceding paragraph, that an antisymmetric tensor could
be represented geometrically by a certain vector w. We will see in Sec-
tion 10-4 how to represent a symmetric tensor. Since antisymmetric and
symmetric tensors have rather different geometric properties, tensors
which occur in physics are usually either symmetric or antisymmetric
rather than a combination of the two. In three-dimensional space, the
introduction of an antisymmetric tensor can always be avoided by the
use of the associated vector. It is therefore not a coincidence that the two
principal examples of tensors in this chapter, the inertia tensor and the
stress tensor, are both symmetric.
10-3 Coordinate transformations. We saw in the previous section that
a tensor T may be defined geometrically as a linear vector operator by
specifying the result of applying T to any vector C. Alternatively, the
10-3] COORDINATE TRANSFORMATIONS 415
tensor may be specified algebraically by giving its components Tij. A
discrepancy exists between the two definitions of a tensor, in that the
algebraic definition appears to depend upon the choice of a particular co-
ordinate system. A similar discrepancy in the case of a vector was noted
in Section 3-1. We will now remove the discrepancy by learning how to
transform the components of vectors and tensors when the coordinate
system is changed. We will restrict the discussion to rectangular co-
ordinates.
Let us consider two coordinate systems, x\, x 2l x 3 , and x[, x' 2 , x 3 , hav-
ing the same origin. The coordinates of a point in the two systems are
related by Eqs. (7-13) :
3
x'i = 23 GijXj, (10-67)
3=1
where
aij = e^- • e,- (10-68)
is the cosine of the angle between the x<- and a^-axes. Likewise,
3
Xj = ^2 a H x 'i- (10-69)
The relations between the primed and unprimed components of any vector
c = X) c'm = E c>i ( 10 - 7 °)
i=l i=l
may be obtained in a similar manner by dotting ej or e^- into Eq. (10-70) :
C't = X) a <A. ( 10 " 71 )
3
i=l
Cj = 2 and. (10-72)
i=l
We can now define a vector algebraically as a set of three components
(Ci, C 2 , C 3 ) which transform like the coordinates {x u x 2 , x 3 ) when the co-
ordinate system is changed. By referring to all coordinate systems, this
definition avoids giving preferential treatment to any particular coordinate
system. In the same way, the primed and unprimed components of a
tensor
T = J2 T '«*& = £ T i*>+ 1 (10 " 73)
»,*=! 3.1=1
416 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
are related by [see Eq. (10-36)]
3
T'i k = el • T • e' k = £ a^cuiT,,, (10-74)
3,1=1
3
T jt = e r T-e,= £ a i} a k iT' ik . (10-75)
i,fc=l
A tensor may be defined algebraically as a set of nine components (Tji)
that transform according to the rule given in Eqs. (10-74) and (10-75).
Note the distinction between a tensor and a matrix. The concept of a
matrix is purely mathematical; matrices are arrays of numbers which may
be added and multiplied according to the rules (10-45) and (10-47). The
concept of a tensor is geometrical; a tensor may be represented in any
particular coordinate system by a matrix, but the matrix must be trans-
formed according to a definite rule if the coordinate system is changed.
The coefficients a,-y defined by Eq. (10-68) are the components of the
unit vectors e* in the unprimed system, and conversely:
3
e'i = X) a t fii, (10-76)
and
-i
£ a v*'i- ( 1( >-77)
Since e[, e' 2t e£ are a set of perpendicular unit vectors, we see that the
numbers 0,7 must satisfy the equations:
3
e » • e * = 23 a « afc > = Sik > (10-78)
3=1
where 5,* is a shorthand notation for
l« = (° *'**' (10-79)
U if i = k.
There are six relations (10-78) among the nine coefficients a ik . Hence,
if three of the constants 0,7 are specified, the rest may be determined from
Eqs. (10-78). It is clear that three independent constants must be specified
to locate the primed axes relative to the unprimed (or vice versa). For
the x[-bx\s may point in any direction and two coordinates are there-
fore required to locate it. Once the position of the a^-axis is determined,
10-3] COORDINATE TRANSFORMATIONS 417
the position of the Xg-axis, which may be anywhere in a plane perpendicu-
lar to x[, may be specified by one coordinate. The position of the Zg-axis
is then determined (except for sign). We can write additional relations
between the o,-/s by the use of such relations as
ej • e* = 5ji, e\ x e' 2 = ±e' 3) ei • (e 2 X e 3 ) = ±1, etc.
(10-80)
Since at least three of the a,/s must be independent, it is clear that the
relations obtained from Eqs. (10-80) are not independent but could be
obtained algebraically from Eqs. (10-78). An interesting relation is ob-
tained from
an <*2i <*3i
e x • (e 2 X e 3 ) =
a 12 a 22 a 32
«13 *23 «33
= ±1, (10-81)
where the result is +1 if the primed and unprimed systems are both
right- or both left-handed and is —1 if one is right-handed and the other
left-handed. Hence the determinant |a,y[ is +1 or —1 according to whether
the handedness of the coordinate system is or is not changed.
In a left-handed system, the cross product is to be defined using the left in
place of the right hand. [In Eq. (10-81), the triple product on the left is to be
evaluated in the primed system.] The algebraic definition is then the same
in either case:
(A X B) = (A 2 B 3 - A3B2, A3B1 - A1B3, A1B2 - A2B1). (10-82)
This definition implies that the cross product A X B of two ordinary vectors is
not itself an ordinary vector, since its direction reverses when we change the
handedness of the coordinate system. An ordinary vector that has a direction
independent of the coordinate system is called a polar vector. A vector whose
sense depends upon the handedness of the coordinate system is called an axial
vector or pseudovector. The angular velocity vector to is an axial vector, and so
is any other vector whose sense is defined by a "right-hand rule." The vector
associated with an (ordinary) antisymmetric tensor is an axial vector. The cross
product w X C of an axial with a polar vector is itself a polar vector. The dis-
tinction between axial and polar vectors arises only if we wish to consider both
right- and left-handed coordinate systems. In the applications in this book, we
need only consider rotations of the coordinate system. Since rotations do not
change the handedness of the system, we shall not be concerned with this dis-
tinction.
418 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
The transformation denned by Eqs. (10-67), (10-71), and (10-74),
where the coefficients satisfy Eq. (10-78), is called orthogonal. As the name
implies, an orthogonal transformation enables us to change from one set
of perpendicular unit vectors to another.
The right member of Eq. (10-71) is formally similar to the right mem-
ber of Eq. (10-34). This suggests an alternative interpretation of Eqs.
(10-71). Let us define a tensor A with components
Atj = ay, (10-83)
and consider the vector
C = A • C. (10-84)
The components C- of C are given by Eq. (10-71). Similarly, by Eq.
(10-72),
C = A' • C. (10-85)
Thus Eqs. (10-71) and (10-72) may be interpreted alternatively as repre-
senting the result of operating with the tensors A, A' upon the vectors C,
C, respectively. In the original interpretation, C ., C'i are components of
the same vector C in two different coordinate systems. In the alternative
interpretation, Cj, C\ are the components of two different vectors C, C
in the same coordinate system. We are primarily interested in the first
interpretation, in which these equations represent a coordinate trans-
formation. However, the latter interpretation will often be useful in
deriving certain algebraic properties of Eqs. (10-71), (10-72), which, of
course, are independent of how we choose to interpret them. In case the
primed axes are fixed in a rotating rigid body, either interpretation is use-
ful. If the primed axes initially coincide with the unprimed axes, then we
may interpret Eqs. (10-71), (10-72) as expressing the transformation from
one coordinate system to the other. Alternatively, we may interpret A
as the tensor which represents the operation of rotating the body from
its initial to its present position, i.e., a vector fixed in the body and ini-
tially coinciding with C will be rotated so as to coincide with C = A • C.
Making use of Eqs. (10-84), (10-85), and (10-43), we deduce that,
A* • (A • C) = (A* • A) ■ C = C. (10-86)
Hence, by Eq. (10-22),
and similarly
A'-A = 1, (10-87)
A A ( = I. (10-88)
A tensor having this property is said to be orthogonal. Equation (10-87)
is evidently equivalent to Eq. (10-78). In the second interpretation, Eqs.
(10-74) and (10-75) can be written as
10-3] COORDINATE TRANSFORMATIONS 419
T' = A • T • A', (10-89)
T = A' • T' • A. (10-90)
The orthogonal tensor is the only example we shall have of a tensor with
a definite geometrical significance, which is neither symmetric nor anti-
symmetric ; it has, instead, the orthogonality property given by Eq. (10-87) .
In view of the fact that the various vector operations were defined with-
out reference to a coordinate system, it is clear that all algebraic rules for
computing sums, products, transposes, etc., of vectors and tensors will be
unaffected by an orthogonal transformation of coordinates. Thus, for
example,
(B + C)' y = B'i + d, (10-91)
(T-C)}= t y 5A ( 10 - 92 )
1=1
<J% = T' H . (10-93)
We can also verify directly the above equations, and others like them,
by using the transformation equations and the rules of vector and tensor
algebra. This is most easily done by taking advantage of the second inter-
pretation of the transformation equations. For example, we can prove
Eq. (10-93) by noting that
(T*)' = A • T' • A'
[by Eq. (10-89)]
= A.(A-T)<
[by Eqs. (10-56) and (10-57)]
= [(A • T) ■ A*]'
[by Eqs. (10-56) and (10-57)]
= (T')',Q.E.D.
[by Eq. (10-89)].
Any property or relation between vectors and tensors which is expressed
in the same algebraic form in all coordinate systems has a geometrical
meaning independent of the coordinate system and is called an invariant
property or relation.
Given a tensor T, we may define a scalar quantity called the trace of T
as follows:
tr(T) = X) T »- ( 10 - 94 )
Since this definition is in terms of components, we must show that the
trace of T is the same in all coordinate systems. In the primed system,
we have
420 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
tr(T) = J2 T '«
1=1
3 3
= Z £ *U*nTji [by Eq. (10-74)]
= Z 2 aijO-u \Tji [rearranging sums]
3
= J2 T H h i l [as in Eq. (10-78)]
}\i=i
= J2 T »> Q- ED - [ b y Ec l- (10-79)].
Another invariant scalar quantity associated with a tensor is the deter-
minant
Tn T12 T13
det(T) = T ai T 22 T 23 , (10-95)
Tzi ^32 ^33
as may also be verified by direct computation.
Let us now study the result of carrying out two coordinate transforma-
tions in succession. The primed coordinates are defined by Eq. (10-67),
in terms of the unprimed coordinates. Let double-primed coordinates be
defined by
x'i = Z a *i x 'i- ( 10 " 96 )
We substitute for x' t from Eq. (10-67) to obtain the double-primed co-
ordinates in terms of the unprimed coordinates:
3 3
X]c = / , / , djciflijXi
= E E a '^ a a \ x i
i=l L i=l J
= 5>jfcry, (10-97)
y-i
where the coefficients of the transformation x — » x" are given by
3
a'kj = X) °W ( 1Q - 98 )
10-4] DIAGONALIZATION OF A SYMMETRIC TENSOR 421
Thus the matrix of coefficients a'^- is obtained by multiplying the matrices
a 'ki> a a according to the rule for matrix multiplication. If we interpret the
transformation coefficients as the components of tensors A, A', A", we
then see from Eqs. (10-45) and (10-98) that
A" = A' • A. (10-99)
This result also follows immediately from Eq. (10-84), applied twice, and
we therefore have an alternative way to derive Eq. (10-98).
10-4 Diagonalization of a symmetric tensor. The constant tensor,
defined by Eq. (10-23), has in every coordinate system* the matrix:
(10-100)
A nonconstant tensor may, in a particular coordinate system, have the
matrix:
/J 1 . \
T = 1 T 2 I • (10-101)
\0 tJ
The tensor T is then said to be in diagonal form. We do not call T a diagonal
tensor, because the property (10-101) applies only to a particular co-
ordinate system; after a change of coordinates [Eq. (10-74)], T will usually
no longer be in diagonal form. If T is in diagonal form, then its effect on
a vector is given simply by
(T • C)< = T&, i = 1, 2, 3. (10-102)
The importance of the diagonal form lies in the following fundamental
theorem:
Any symmetric tensor can be brought into diagonal form by
an orthogonal transformation. The diagonal elements are then
unique except for their order, and the corresponding axes are
unique except for degeneracy. (10-103)
Before proving this important theorem, let us try to understand its
significance. The theorem states that, given any symmetric tensor T,
we can always choose the coordinate axes so that T is represented by a
diagonal matrix. Furthermore, this can be done in essentially only one
way; there is only one diagonal form (10-101) for a given tensor T, except
: See Problem 10 at the end of this chapter.
422 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
for the order in which the diagonal elements T\, T 2 , T 3 appear, and each
element is associated with a unique axis in space, except for degeneracy,
that is, except when two or three of the diagonal elements are equal. The
axes ei, e 2 , e3 in the coordinate system in which the tensor has a diagonal
form are called its principal axes. The diagonal elements T\, T 2 , T% are
called the eigenvalues or characteristic values of T. Whenever we write a
tensor element with a single subscript, we shall mean it to be an eigenvalue.
Any vector C parallel to a principal axis is called an eigenvector of T.
An eigenvector, according to Eq. (10-102), has the property that opera-
tion by T reduces to multiplication by the corresponding eigenvalue:
T • C = TiC, (10-104)
where Ti is the eigenvalue associated with the principal axis e» parallel
toC.
The theorem (10-103) allows us to picture a symmetric tensor T as a
set of three numbers attached to three definite directions in space. If
we think of T as applied to each vector C in the vector space, then formula
(10-102) shows that the effect is a stretching or a compression along each
principal axis, together with a reflection if T,- is negative. In Section 10-2
we saw that a symmetric tensor may be specified by giving six components
Tij in any arbitrarily chosen coordinate system. We now see that we can
alternatively specify T by specifying the principal axes (this requires three
numbers, as we have seen), and the three associated eigenvalues.
If two or three of the eigenvalues are equal, we say the eigenvalue is
doubly or triply degenerate. If the eigenvalue is triply degenerate, the
tensor clearly is a constant tensor [Eq. (10-100)] and diagonal in every
coordinate system. The principal axes are no longer unique; any axis
is a principal axis. Every vector is an eigenvector of a constant tensor.
If two eigenvalues are equal, say T x — T 2> then if we consider rotating
the coordinate axes in the eie 2 -plane, we can see that the tensor will re-
main in diagonal form; the four elements referring to this plane behave
like a constant tensor in that plane. Again the principal axes are not
unique, since two of them may lie anywhere in the eie2-plane. The third
axis e3 associated with the nondegenerate eigenvalue T3, however, is
unique. We can prove that every axis in the e^-plane is a principal
axis by considering the effect of T on any vector
C = dd + C 2 e 2 (10-105)
in this plane. In view of Eq. (10-102), if T x = T 2 , we have
T • C = Tide! + T 2 C 2 e 2
= TiC, (10-106)
10-4]
DIAGONALIZATION OF A SYMMETRIC TENSOR
423
so that C is an eigenvector of T. Every vector in the eie 2 -plane is an
eigenvector of T with eigenvalue T t . If we like, we may say that there
is a principal plane associated with a doubly degenerate eigenvalue.
We will now prove the theorem (10-103) by showing how the principal
axes can be found. Let a symmetric tensor T be given in terms of its com-
ponents Tn in some coordinate system, which we will call the initial co-
ordinate system. To find a principal axis, we must look for an eigenvector
of T. Let C be such an eigenvector, and T' the corresponding eigenvalue.
We can rewrite Eq. (10-104) in the form
(T — T'l)-C = 0. (10-107)
If we write this equation in terms of components, we obtain
(fii - T')C l + T 12 C 2 + T 13 C 3 = 0,
TaiCi + (T 22 - T')C 2 + T 23 C 3 = 0, (10-108)
TnCi + T 32 C 2 + (T 33 - T')C 3 = 0.
These equations for the unknown vector C have, of course, the trivial
solution C = 0. If we write the solution for C,- in terms of determinants,
we see that C = is the only solution unless the determinant
Tu - T' T 12 T l3
rrp rpt rp
21 * 22 — J- l 23
rp rp rp rpt
1 31 *■ 32 i 33 — 1
o,
(10-109)
in which case the solution for C,- is indeterminate. In this case it is shown
in the theory of linear equations* that Eqs. (10-108) have also nontrivial
solutions d. It is clear that Eqs. (10-108) cannot determine the numbers
Ci, C 2 , C 3 uniquely, but only their ratios to one another, Ci:C 2 :C 3 . This
is also clear from Eq. (10-107), from which we began. Geometrically,
only the direction of C is determined, not its magnitude (nor its sense).
Equation (10-109), called the secular equation, represents a cubic equation
to be solved for the eigenvalue T'. In general there will be three roots
T[, T 2 , T' 3 . Given any root T', we can then substitute it in Eqs. (10-108)
and solve for the ratios Ci:C 2 '-C 3 . Any vector whose components are in
the ratio C\:C 2 :C 3 is an eigenvector of T corresponding to the eigenvalue
T'. For each eigenvalue T' it we can then take a unit vector e'j along the
direction of the corresponding eigenvectors. The axes e[, e' 2> e 3 are then
* See, for example, Knebelman and Thomas, Principles of College Algebra.
New York: Prentice-Hall, Inc., 1942. (Chapter IX, Theorem 10.)
424 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
the principal axes of T. When we solve Eqs. (10-108) for the components
of e'j for a given T' jt we get three numbers ay,- (=C» for T' = T' } ) which
are the components of e^ along the axes e,- of the initial coordinate system:
e'y = E *&*■ (10-110)
»=i
This is just Eq. (10-76), hence the numbers a# are the coefficients of the
orthogonal transformation from the initial coordinate system to the prin-
cipal axes. We say that the transformation with coefficients a,i diago-
nalizes T.
In order to be sure we can carry out the above program, we must prove
three lemmas, as the reader may have noted. First, we must prove that
the roots T' of the secular equation (10-109) are real; otherwise we cannot
find real solutions of Eqs. (10-108) for C u C 2 , C 3 . Second, we must prove
that the vectors ej, obtained from Eqs. (10-108) for the different eigen-
values T'j, are perpendicular; otherwise we do not obtain a set of perpen-
dicular unit vectors. Third, we must show that in the degenerate case,
two (or three) perpendicular unit vectors ej can be found that correspond
to a doubly (or triply) degenerate eigenvalue.
Lemma 1. The roots of the secular equation (10-109) for a
symmetric tensor are real. (10-111)
Equation (10-109) is obtained from the eigenvalue equation (10-104) :
T • C = T'C. (10-112)
To prove the lemma, let us first allow T' to be complex. We will need
also to allow the components C» of the vector C to be complex. A vector
C with complex components, has no geometric meaning in the usual sense,
of course, but we can regard all the algebraic definitions of the various
vector operations as applying also to vectors with complex components.
The various theorems of vector algebra will hold also for vectors with
complex components. [There is one exception to these statements. The
length of a complex vector cannot be defined by Eq. (3-13), but instead
must be defined by
|A| = (A* • A)*. (10-113)
This definition will not be required here.] We will denote by C* the vector
whose components are the complex conjugates of those of C. Let us multi-
ply C* into Eq. (10-112):
C* • T • C = 2T(C* • C). (10-114)
10-4] DIAGONALIZATION OF A SYMMETRIC TENSOR 425
If we take the complex conjugate of this equation, we have
C • T • C* = r*(C* • C), (10-115)
since T is real, and in view of Eq. (3-18). Now by definition (10-53),
C* • T = T' • C*. (10-116)
Hence
C* • T • C = (C* • T) • C
= (¥' • C*) • C [by Eq. (10-53)]
= C • T' • C* [by Eq. (3-18)]. (10-117)
For a symmetric tensor, T = T', so that the left members of Eqs. (10-114)
and (10-115) are equal, and
T' = T'*, (10-118)
so that T' is real.
Lemma 2. The eigenvectors of a symmetric tensor correspond-
ing to different eigenvalues are perpendicular. (10-119)
To prove this lemma, let us assume that T[, T 2 are two eigenvalues of T
corresponding to the eigenvectors C 1( C 2 :
T Ci = TiC 1( (10-120)
T C 2 = r 2 C 2 . (10-121)
We multiply C 2 into Eq. (10-120), and Ci into Eq. (10-121):
C 2 • T • Ci = 2"i(C a • Ci), (10-122)
d • T • C 2 = T' 2 (C 2 • Ci). (10-123)
Since T is symmetric, the left members are equal, and we have
(Ti - r'a)(C a • d) = 0. (10-124)
If the eigenvalues T[, T' 2 are unequal, the eigenvectors C 2 , C x are per-
pendicular.
Lemma 3. In the case of double or triple degeneracy, Eqs.
(10-108) have two or three mutually perpendicular solutions
for the vector C. (10-125)
For the proof of (10-125), suppose that T[ = T 2 . If we substitute
T' = T[ in Eqs. (10-108), then by the theorem referred to in the
426
TENSOR ALGEBRA. INERTIA AND STRESS TENSORS
[CHAP. 10
footnote on page 423, there is at least one nontrivial solution C\, C2,
C 3 . Let e( be a unit vector parallel to the vector (C\, C 2 , C 3 ). Then
T • e' x = TWi.
(10-126)
Now, choose any pair of perpendicular unit vectors e 2 ' , e 3 ' perpendicular
to ej, and use Eqs. (10-68) and (10-74) to transform the components of
T into the double-primed coordinate system e^ , e 2 ', e 3 '. By a comparison
of Eqs. (10-126) and (10-36), we see that we must get
T'1'1 = T' u Till = 0, r'3'1
0.
(10-127)
Since T is symmetric, its double-primed components must therefore be
given by
(T' x \
T = I T' 2 ' 2 r'2'3 • (10-128)
\o r 2 ' 3 T'J
Furthermore, the secular equation
T\ - T'
T'2'2 - T' r'2'3
= (10-129)
T' 2 ' 3 n's - T'
must have the same roots as Eq. (10-109). This is true since the left mem-
bers of both equations are the determinants of the same tensor (T — T'l),
expressed in the unprimed and double-primed coordinate systems, and we
noted at the end of Section 10-3 that the determinant of a tensor has the
same value in all coordinate systems. If we expand the determinant
(10-129) by minors of the first row, we obtain
(Ti - T')
22
T
T'l
23
23
n'3 - r
= 0.
(10-130)
Since T[ is a double or triple root of this equation, it must be a root of the
equation
rpil TV mil
(10-131)
T'2'2 - T' T' 2 ' 3
= 0.
rpn rptt rpt
J- 23 * 33 — ■*■
Therefore the equations
(T'2'2 - Ti)C' 2 ' + TWi = 0,
T
V'hC'i + (T'3'3 - T'x)C
{ = 0,
(10-132)
10-4]
DIAGONALIZATION OF A SYMMETRIC TENSOR
427
have a nontrivial solution which defines an eigenvector (0, C 2 , C 3 ) in the
e^'eg -plane w i tn tne eigenvalue T[. We have therefore a second unit
eigenvector e 2 parallel to (0, C 2 , C 3 ) and perpendicular to e[. If we take
a third unit vector e^ perpendicular to e[, e 2 , then in this primed coor-
dinate system, we must have
T'u = T lt T' 21 = 0,
T' 12 = 0, T' 22 = n,
Thus T must have the components
Tii = 0,
(10-133)
32
T =
fTi >
Ti
\0 n/
(10-134)
and e{, e 2 , e^ are principal axes. If T[ were a triple root of Eq. (10-109),
it would also be a triple root of the secular equation
Ti
T'i
V
(Ti - T'XTi - T')(T 3 - T') = 0.
(10-135)
Therefore T' 3 = T[, and we have three perpendicular eigenvectors cor-
responding to the triple root T[ = T 2 = T' 3 .
The above three lemmas complete the proof of the fundamental theorem
(10-103).
The algebra in this section may be generalized to vector spaces of any
number of dimensions, with analogous results regarding the existence of
principal axes of a symmetric tensor.
At the end of Section 10-3 we noted that the trace and the determinant of
a tensor T have the same value in all coordinate systems. We see from Eq.
(10-101) that the trace is the sum of the eigenvalues of T:
tr(T) = T x + T 2 + T 3>
and the determinant is the product of the eigenvalues:
det(T) = T1T2T3.
(10-136)
(10-137)
We can form a third invariant scalar quantity associated with a symmetric
tensor by summing the products of pairs of eigenvalues :
M(X) = TiT 2 + T 2 T 3 + T 3 Tl
(10-138)
428
TENSOR ALGEBRA. INERTIA AND STRESS TENSORS
[CHAP. 10
We can evaluate M (T) in any coordinate system by solving the secular equation
(10-109) for the three roots T x , T 2 , T 3 and using Eq. (10-138). The solution of
Eq. (10-109) can be avoided by noting that the sum (10-138) must be the
coefficient of T' in Eq. (10-109), which is the sum of the diagonal minors of the
determinant of T :
T33T31
M(T) =
TuT 12
T21T22
+
T22T23
T32T33
+
T13T11
(10-139)
We could also show by direct calculation that M(T) as given by Eq. (10-139)
has the same value after a coordinate transformation given by Eq. (10-74).
For any tensor T, the determinant of (T — T'\) must have the same value
in all coordinate systems. Therefore, in particular, the roots T' of Eq. (10-109)
will be the same in all coordinate systems, even for a tensor T that is not sym-
metric. We still call the roots T' the eigenvalues of T. If T is not symmetric,
one eigenvalue will be real and the other two will be a conjugate complex pair.
For the real eigenvalue we can find an eigenvector. For the complex eigen-
values we cannot, in general, find eigenvectors. (That is, not unless we admit
vectors with complex components, which have only algebraic significance. Even
then, we cannot prove in general that the eigenvectors are orthogonal.) In any
case, the expressions given by Eqs. (10-136), (10-137), (10-138), and (10-139)
are still real and independent of the coordinate system.
As an example of the diagonalization procedure, let us diagonalize the
tensor
T = AA + BD + DB,
which obviously is symmetric. We will take
A = 4aei,
B = 7ae 2 + ae 3 ,
D = ae2 — 0,63.
The tensor T is then represented in this coordinate system by the matrix:
T =
14a 2
-6a' —2a'
In this case, the secular equation (10-109) is
16a 2 - T
14a 2 - T
-6a 2
-6a 2
-2a 2 - T
= (16a 2
T')
X (T' 2 + 12a 2 T' - 64a 4 ) = 0.
10-4] DIAGONALIZATION OF A SYMMETRIC TENSOR 429
The roots (necessarily real) are
T\ = 16a 2 , T' 2 = 16a 2 , T' 3 = -4a 2 .
Equations (10-108), for the doubly degenerate root T' = 16a 2 , are
0=0,
-2a 2 C 2 - 6a 2 C 3 = 0,
-6a 2 C 2 - 18a 2 C 3 = 0.
Clearly, Ci is arbitrary, and the last two equations are both satisfied if
C/2 == — 3t7 3 .
Therefore any vector of the form
C = Ciei - 3C 3 e 2 + C 3 e 3
is an eigenvector for arbitrary C\, C 3 . Thus we have a two-parameter
family of possible eigenvectors, from which we may select for e[, e 2
any two perpendicular unit vectors. We will take
ei = ei,
,31
©2 — — ;= 62 7= e 3 .
Vio vlo
We could have guessed e[ from the form of T. The reader should verify
that ej and e 2 and, in fact, any vector in the eie2-plane satisfy Eq.
(10-104) with T{ = 16a 2 . For T' = —4a 2 , Eqs. (10-108) become
20a 2 C! = 0,
18a 2 C 2 - 6a 2 C 3 = 0,
-6a 2 C 2 + 2a 2 C 3 = 0.
Now there is just a one-parameter family of solutions
C\ = 0, C 3 = 3C 2 ,
of which there is one unit eigenvector (except for sign) :
1 , 3
e 3 = . — e 2 -\ — e 3 ,
Vio Vio
where positive signs were chosen so that ej, e 2 , e^ would form a right-
handed system. The vector e^ is perpendicular to the eje 2 -plane as it
430 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
must be according to lemma 1. It may be verified that e 3 is an eigen-
vector of T with the eigenvalue —4a 2 .
By reference to Eq. (10-76), we may write the coefficients of the trans-
formation to the principal axes of T:
(an ai2 ai3\ A ° ° \
a 2 i a 22 a 23 )=lo 3/VW -l/VIo).
a 3 i o 32 a 33 / \0 1/VlO 3/VlO /
The reader should verify that these coefficients satisfy Eqs. (10-78) ; that
the vectors ej are properly transformed according to Eqs. (10-71) and
(10-72), which in this case are
3 3
Sjfc = 2~i °*»*jF»> ^a = 2^i aki ^'*>
where e£ is the rth component of e'j in the unprimed coordinate system
and 8jk is the fcth component of ej in the primed coordinate system; and
also that T is properly transformed according to Eq. (10-74) from its
original form to its diagonal form.
10-5 The inertia tensor. The inertia tensor of a rigid body is given by
Eq. (10-25). For a body of density p(x, y, z), we may rewrite the inertia
tensor as ,,,
| = jjfp{r 2 \ - it) dV, (10-140)
where we have used the subscript "o" to remind us that the inertia tensor
is calculated with respect to a set of axes with origin at O. We will omit the
subscript except when the discussion concerns more than one origin. The
diagonal components of I are just the moments of inertia [Eq. (5-80)]
about the three axes:
l xx = fff P (y 2 + z 2 ) dV,
Iyy = ///P(* 2 + x 2 ) dV, (10-141)
hz = fffp(x 2 + y 2 ) dV.
The off-diagonal components, often called products of inertia, are
Ixy = Iyx= — fffpxy dV >
I V z ==/.„=- fffpyz dV, (10-142)
Izx = Ixz = — fffP ZX dV -
10-5]
THE INERTIA TENSOR
431
Since we may use Eq. (10-74) to calculate the components of the
inertia tensor relative to any other set of axes through the same origin 0,
we see from Eqs. (10-74) and (10-141) that the moment of inertia about
any axis through 0, in a direction designated by the unit vector n, is
J„ = n
n.
(10-143)
It often is easier to calculate the components of the inertia tensor with
respect to a conveniently chosen set of axes and then use Eq. (10-143),
than to calculate 7 n directly, if the axis n is not an axis of symmetry of the
body.
We can obtain a useful analog to the Parallel Axis Theorem (5-81) for
the moment of inertia by calculating the inertia tensor \ relative to an
arbitrary origin of coordinates in terms of the inertia tensor Iff rela-
tive to the center of mass G. Let r and r' be position vectors of any point
P in the body relative to O and G respectively, and let R be the coordinate
of G relative to (Fig. 5-12),
r = r' + R.
(10-144)
Then we have, from Eq. (10-140),
\ = ///p[(r' + R) ■ (f + R)l - Of + R)(r> + R)] dV
= fffplif ■ r')l - r'r'] dV + [(R • R)l - RR]///p dV
+ 21 [R • ///pr' dV] - [fff P r> dV] R - R///pr' dV.
(10-145)
In view of the definition (5-53) of the center of mass, we have
jjfpr' dV = 0. (10-146)
Equation (10-145) therefore reduces to
l„ = l G + M(fl 2 l — RR).
(10-147)
Note that both the statement and proof of this theorem are in precise
analogy with the Parallel Axis Theorem (5-83) for the moment of inertia.
It is evident from the definition (10-140) that the inertia tensor of a
composite body may be obtained by summing the inertia tensors of its
parts, all relative to the same origin.
If a body rotates, the components of its inertia tensor, relative to sta-
tionary axes, will change with time. The components relative to axes
fixed in the body, of course, will not change if the body is rigid. We may
think of the inertia tensor I as rotating with the body. If the (constant)
432 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
components along axes fixed in the body are given, the (changing) com-
ponents along stationary axes are then given by Eq. (10-89), where A
represents the transformation from, body axes to space axes. The most
convenient set of axes in the body for most purposes are the principal
axes of the inertia tensor, also called the principal axes of the body. The
eigenvalues of the inertia tensor are called the principal moments of in-
ertia. We will learn more in the next chapter of the dynamical significance
of the principal axes, but we may note here that, according to Eq. (10-24),
if the body rotates about a principal axis, the angular momentum is paral-
lel to the angular velocity. We may always choose arbitrary axes, compute
I, and then use the method of Section 1-4 to find the principal axes. It is
often possible, however, to simplify the problem by choosing to begin with
a coordinate system in which one or all of the axes are principal axes.
In many cases, a body will have some symmetry, so that we can see
that certain of the products of inertia (10-142) will vanish if the axes are
chosen in a certain way. For example, we can prove the following theorem:
Any plane of symmetry of a body is perpendicular to a prin-
cipal axis. (10-148)
If we choose the yz-plane as the plane of symmetry, then
p(-x, y, z) = p(x, y, z). (10-149)
It is easy to show that because of Eq. (10-149) the integrals (10-142) for
I X y and I iX will vanish. Therefore the a;-axis is a principal axis in this case.
In a similar way, we can prove the theorem:
Any axis of symmetry of a body is a principal axis. The plane
perpendicular to this axis is a principal plane corresponding
to a degenerate principal moment of inertia. (10-150)
A sphere, or a body with spherical symmetry has, evidently, a constant
inertia tensor.
As an example, consider the right triangular pyramid shown in Fig.
10-1. The components of the inertia tensor relative to the axes (x, y, z)
are to be calculated from the formula:
/•Jo ra-%t j.a-y-%z /y 2 -\- z 2 — X y —ZX \
I = / / / pi — xy z 2 + x 2 —yz I dx dy dz,
J z=0 J y=0 Jx=0 \ 2 I 2/
\ — zx —yz x + y I
where each component of I is to be obtained by evaluating the indicated
integral over the corresponding component of the matrix. The density
p is given in terms of the mass M by
M = i a 3 p.
10-5]
THE INERTIA TENSOR
433
Fig. 10-1. A right triangular pyramid.
Because of the symmetry between x and y, it is necessary to evalute only
the four integrals
J i = fllPZ 2 dx dy dz = [[[py 2 dx dy dz = -fa Ma 2 ,
J 2 = [I I pz 2 dx dydz = ^g Ma 2 ,
Jz = lijpxy dx dydz = -g$ Ma 2 ,
J 4 = III pxz dx dy dz = Iffpyz dx dy dz = ^ Ma 2 .
The inertia tensor is then given by
(Ji + J 2 -Js -J*\ I 13 -2 -3)
:;)■(:
Ji + J 2 -J 4 ) = \ -2
-J 4 2JJ \-3
where the notation means that each element of the matrix is to be mul-
tiplied by Ma 2 /40. Let us find the principal axes. By symmetry [theorem
(10-148)] the axis x" shown in Fig. 10-1 is a principal axis. Let us there-
fore first transform to the axes x", y", z. The coefficients of the trans-
formation are, by Eq. (10-68),
(ax», a x - v a x »\ jl/y/2 -1/V2 0\
o»»« ay. y <V'J=(l/\/2 1/V2 0l-
a-zx o. zy a zl / \0 1/
434
TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
Using Eq. (10-74), we now calculate the inertia tensor components along
the x"-, y"-, and z-axes:
I
r 15
11
>
-3V2
Ma 2
40
\ -3V2 8
We see that the x"-axis is indeed a principal axis. The secular equation is
15 - X
11 - X -3V2 =0, T' =
-3V2 8 - X
and the roots are
\ x > = 15, \ v ' = 5, X z < = 14,
Ma'
40
X,
or
2V = £ Ma 2 , T y . = i Ma 2 , TV = £> Ma 2
Equations (10-108) can be solved for the components of the unit vectors
i', j', k' in terms of i", j", k:
i' = i",
j' = 4V3J" + *V6k,
k' = -fVoj" + iVSk.
As a second example, let us find the inertia tensor about the point of
the object shown in Fig. 10-2. The object is composed of three flat disks
Fig. 10-2. Three disks.
Fig. 10-3. A circular disk with its
principal axes.
10-5] THE INERTIA TENSOR 435
of mass M and radius a. By symmetry, the principal axes are the indi-
cated axes x, y, z. We first calculate the inertia tensor of a single disk
about its center, relative to its principal axes x', y', z', as shown in Fig.
10-3. The moment of inertia I z ' is given by Eq. (5-90), and the moments
of inertia I x >, I y - are half I z >, by the Perpendicular Axis Theorem (5-84).
We can therefore write the inertia tensor of a disk, relative to its principal
axes x', y', z', as
/l 0\ „, 2
(10-151)
For the bottom disk, the principal axes are parallel to x, y, z, and we need
only apply theorem (10-147) to obtain its inertia tensor relative to the
x-, y-, z-axes with origin at O:
l = l ff + M(3a 2 l - 3a 2 kk)
/l3 0\ ,, 2
= o 13 )*%-.
\0 2/ 4
For the right-hand disk, with the axes x', y', z' oriented as shown, we first
apply theorem (10-147) to obtain the inertia tensor about 0, relative to
axes parallel to x', y', z'\
(b 0\ M 2
l.cv.')=(0 10^-
\0 6/
The transformation from x'-, y'-, z'-axes to x-, y-, z-axes is given by
(a xx > a xy ' a xz \ /l \
a yx ' a vv > a yz > I = I J \y/Z I •
a zx ' a zy > a zz >l \0 —\\/Z J /
We now use Eq. (10-74). It is perhaps easier to carry out the process in
two steps, according to Eq. (10-89):*
(5 \ „ 2
i 3V3)^-
-i\/3 3 /
* Matrices may be multiplied conveniently according to the rule (10-45)
by noting that the element (T • S)i* is obtained by summing the products of
pairs of elements across row i in T and down column k in S.
436 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
Now
\/l
v/3
(A • l.(,v.',) • A' = I i 3V3 1 1 J
\0 -iV3 3 /\0 iV3
= ■.<*,.) = 10 ^ |V3) Ma
2
We may interpret this algebra as a computation of l relative to a new
set of axes. Alternatively, we may interpret A as a tensor which rotates
the disk through an angle of 60° about the z-axis; l (xv*') * s * nen tne
moment of inertia of a disk whose principal axes are parallel to x, y, z, and
the algebra is a computation of the effect of rotating the disk to its final
position. The left-hand disk, correspondingly, has the inertia tensor
\ 7M- 2
V -*V5)^-
-iV3
l
The inertia tensors of the three disks may now be added and we obtain
Ma 2
f23 0\ „ 2
l = I 22J
\0 6i/ '
Let us calculate the moment of inertia of the object shown in Fig. 10-2
about the y'-axis through 0. By Eq. (10-143), we have
h = J' • U • j' = lOpfa 2 .
We could use theorem (10-147) to obtain I about the center of gravity G,
which is at the intersection of the z- and z'-axes. It is clear from symmetry
that any axis perpendicular to one of the three disks through its center is
a principal axis relative to G. This can only be true if the inertia tensor rel-
ative to G has a double degeneracy in the yzy'z' '-plane. The reader should
check this by carrying out the translation of l„ to the center of mass G.
It should be noted that the principal axes of the inertia tensors of a body
relative to two different points and 0', in general, will not be parallel,
as experimentation with Eq. (10-147) will show.
The kinetic energy T of a rotating rigid body can also be expressed con-
veniently in terms of the inertia tensor. From Eqs. (10-2) and (10-3)
and using the rules of vector algebra, we have
10-5]
THE INERTIA TENSOR
437
TV
T = J2 W*
N
N
7c=l
= i« • l.
Therefore T can be expressed in the form
T = \u> ■ I • w.
(10-152)
(10-153)
Equation (10-153) expressed in terms of components along any set of
axes is then
(10-154)
This is the equation of a family of quadric surfaces in w-space, each sur-
face the locus of angular velocities for which the kinetic energy has a con-
stant value T. If Eq. (10-153) is written in terms of components along
principal axes x', y', z',
w; + \iw + wt = t,
i r',,' 2 —
(10-155)
then we see that these surfaces are ellipsoids, since the moments of inertia
are necessarily positive. If we define a vector
_ a
where a is a constant, then Eq. (10-153) can be written as
r • I • r = o
(10-156)
(10-157)
This is the equation of the inertia ellipsoid. The constant a determines the
size of the ellipsoid. It is customary to set a = 1 in whatever units are
being used, for example, a = 1 cm-erg-sec. In this case, we note that the
size of the ellipsoid (but not its shape) depends on the units being used.
The inertia ellipsoid of a body, like its inertia tensor, is relative to a
particular origin about which moments are computed. The six coefficients
of. the quadratic form on the left of Eq. (10-157) are the components of
the inertia tensor:
I xx x 2 + I yv y 2 + I zl z 2 + 2I xy xy + 2I yz yz + 2I tx zx = a 2 , (10-158)
438 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
so that the inertia tensor is uniquely characterized by the corresponding
inertia ellipsoid. This gives us another convenient geometrical way of
picturing the inertia tensor.
By comparing Eq. (10-157) with Eq. (10-143), we see that the radius
to any point on the inertia ellipsoid is
r = air*, (10-159)
where I t is the moment of inertia about an axis parallel to r. In particular,
the principal moments of inertia are related by Eq. (10-159) to the semi-
principal axes of the inertia ellipsoid. We see that if there is double de-
generacy, the inertia ellipsoid is an ellipsoid of revolution. If the principal
moments of inertia are all equal, the ellipsoid of inertia is a sphere.
For any symmetric tensor T, we can form a quadratic equation of the
form (10-157) which defines a quadric surface that uniquely charac-
terizes T. The principal axes of T are the principal axes of its associated
quadric surface. If the eigenvalues of T are all positive, the surface is an
ellipsoid. Otherwise, it will be a hyperboloid or a cylinder. If all the
eigenvalues are negative, we would need to write —a 2 for the right member
of the quadratic equation in order to define a real surface.
10-6 The stress tensor. Let us represent any small surface element in
a continuous medium by a vector dS whose magnitude dS is equal to the
area of the surface element and whose direction is perpendicular to the
surface element. To specify the sense of dS, we will distinguish between
the two sides of the surface element, calling one the back and the other
the front. The sense of dS is then from the back to the front. We may
then describe the state of stress of the medium at any point Q by specify-
ing the force P(dS) exerted across any surface element dS at Q by the
matter at the back on the matter at the front of dS. We understand, of
course, that the surface element dS is infinitesimal. That is, all statements
we make are intended to be correct in the limit when all elements dS — > 0.
For a sufficiently small surface element, the force P may depend on the
area and orientation of the surface element, but not on its shape. Thus P
is indeed a function only of the vector rfS at any particular point Q in the
medium. We will show that P(dS) is a linear function of dS. We may
therefore represent the function P(dS) by a tensor P, the stress tensor:*
P(eZS) = P • dS. (10-160)
* The reader is cautioned that many authors define the stress tensor with the
opposite sign from the definition adopted here, so that a tension is a positive
stress and a pressure, a negative stress. The latter convention is almost universal
in engineering practice, whereas the definition adopted here is more common in
works on theoretical physics.
10-6] THE STRESS TENSOR 439
P(dS 2 )
> P(-dS! — dS 2 )
-(dSj + dS 2 )
Fig. 10-4. A triangular prism in a continuous medium.
To show that P(dS) is a linear vector function, we note first that if dS
is small enough so that the state of stress of the medium does not change
over the surface element, then the force P will be proportional to the
area dS so long as the orientation of the surface is kept fixed. Thus for
a positive constant c,
P(cdS) = cP(dS). (10-161)
If the direction of dS is reversed, the back and front of the surface ele-
ment are interchanged, and therefore by Newton's third law, P(— dS) =
— P(dS), so that Eq. (10-161) holds also if c is negative. Now, given any
two vectors dS x , dS 2 , let us imagine a triangular prism in the medium with
two sides dS u dS 2 , as in Fig. 10-4. If the end faces are perpendicular to
the sides, then the third side is — dSi — dS 2 , as shown. If the length of
the prism is made much greater than the cross-sectional dimensions, we
may neglect the forces on the end faces, and the total force on the prism is
dF = P(dSx) + P(dS 2 ) + P(-dSt - dS 2 ). (10-162)
If the density is p, the acceleration of the prism is given by Newton's
law of motion:
p dVa = dF. (10-163)
Now if we reduce all linear dimensions of the prism by a factor a, the areas
dSi are multiplied by a 2 ; hence by Eq. (10-161), dF is multiplied by a 2 ,
440 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
and dV is multiplied by a 3 , so that
ap dV&(a) = dF, (10-164)
where a(a) is the acceleration of a prism a times smaller. Now as a — * 0,
the acceleration should not become infinite; hence we conclude that
dF = 0, (10-165)
from which, by Eqs. (10-162) and (10-161),
P(dS!) + P(eZS 2 ) = P(dS x + dS a ). (10-166)
Equations (10-161) and (10-166) show that the function P(dS) is linear.
Note that Eqs. (10-166) and (10-162) imply that there is no net force on
the prism if the stress function P(rfS) is the same at all faces. Any net
force can only result from differences in the stress at different points of
the medium; such differences reduce to zero as a — ► 0.
By considering small square prisms, and recognizing that the angular
acceleration must not become infinite as the size shrinks to zero, we can
show by a very similar argument (see Problem 32) that P must be a
symmetric tensor. The stresses at each point Q in a medium are therefore
given by specifying six components of the symmetric stress tensor P.
If the medium is an ideal fluid whose only stress is a pressure p in all
directions, the stress tensor is evidently just
P = pi. (10-167)
Note that we did not prove in Chapter 8 that in an ideal fluid, that is,
one which can support no shearing stress, the pressure is the same in all
directions. This was proved only in Chapter 5 for a fluid in equilibrium.
This logical defect can now be remedied. (See Problem 33.)
According to the definition of P, the total force due to the stress across
any surface S is the vector sum of the forces on its elements:
//'
'P-dS. (10-168)
s
If S is the closed surface surrounding a volume V of the medium, and if
we take n to be the conventional outward normal unit vector, then the
total force exerted on the volume V by the matter outside it is
F = - //n • P dS, (10-169)
s
and by the generalized Gauss' theorem, [see discussion below Eq. (5-178)],
F = - fffv ■ P dV. (10-170)
v
10-6] THE STRESS TENSOR 441
Since V is any volume in the medium, the force density due to stress is
f s = -V P. (10-171)
In agreement with an earlier discussion, we see that this force density
arises only from differences in stress at different points in the medium.
Equation (10-171) may also be derived by summing the forces on a small
rectangular volume element.
The equation of motion (8-138) may now be generalized to apply to any
continuous medium:
P ^ + V ■ P = f. (10-172)
This equation may also be rewritten in the form (8-139) :
J + T .Vv + Jv.P = f (10-173)
at p p
This equation, together with the equation of continuity (8-127), de-
termines the motion of the medium when the body force density f and the
stress tensor P are given. The stress P at any point Q may be a function
of the density and temperature, of the relative positions and velocities
of the elements near Q, and perhaps also of the previous history of the
medium, which may be a solid (elastic or plastic) or a fluid (ideal or
viscous).
From Eqs. (10-172) and (10-173) we can derive conservation equa-
tions analogous to those derived in Section 8-8. The conservation equation
for energy analogous to Eq. (8-149) is, for example,
j t {W 5F) = v • (f - V • P) bV: (10-174)
The further manipulations of the energy equation carried out in Sec-
tion 8-8 cannot all be carried through in the same way for Eq. (10-174)
because of the difference in form between the stress term here and the
pressure term in Eq. (8-149), as the reader may verify. The energy
changes associated with changes in volume and shape of an element in a
continuous medium are in general more complicated than those associated
with expansion and contraction of an ideal fluid.
In a viscous fluid, the stress tensor P will be expected to depend on the
velocity gradients in the fluid. This is consistent with the dimensional
arguments in Section 8-14, where we saw that the term V • P in
Eq. (10-172) must consist of the coefficient of viscosity i\ multiplied by
some combination of second derivatives of the velocity components with
respect to x, y, and z. If the fluid is isotropic, as we shall assume, then
the relation between P and the velocity gradients must not depend on the
orientation of the coordinate system. We can guarantee that this will
442
TENSOR ALGEBRA. INERTIA AND STRESS TENSORS
[CHAP. 10
be so by expressing the relation in a vector form that does not refer ex-
plicitly to components. The dyad
j dv x
dx
Vv =
dVy
dx
dx
f\
dv x dv y dv z
dy dy by
\ dv x
\dz
dVy
dz
dv z
dz
(10-175)
has as its components the nine possible derivatives of the components of
v with respect to x, y, and z. Hence we must try to relate P to Vv.
The dyad (10-175) is not symmetric, but we can separate it into a sym-
metric and an antisymmetric part, as in Eqs. (10-64) through (10-66):
Vv = (W). + (Vv)„,
(Vv). = ivv + i(w)' f
(Vv)„ = JT7- i(Vv)'.
(10-176)
(10-177)
(10-178)
The antisymmetric part is related, as in Eqs. (10-62) and (10-63), to a
vector
«=|VXv, (10-179)
such that for any vector dx,
(Vv)„ • dx = u X dx.
(10-180)
If dx is the vector from a given point Q to any nearby point Q', we see that
the tensor (Vv) a selects out those parts of the velocity differences between
Q and Q' which correspond to a (rigid) rotation of the fluid around Q with
angular velocity a. This is in agreement with the discussion of Eq. (8-133),
which is identical with Eq. (8-179). Since no viscous forces will be asso-
ciated with a pure rotation of the fluid, the viscous forces must be expres-
sible in terms of the tensor (Vv)..
Since P is also symmetric, we are tempted to write simply
P = C(Vv).,
(10-181)
where C is a constant. In the simple case depicted in Fig. 8-10, the only
nonzero component of Vv is dv x /dy, and Eqs. (10-181) and (10-177) then
give
iC% o\
It » o
dy
\%c
(10-182)
.
10-6] THE STRESS TENSOR 443
and the viscous force across dS = j dS will be
dF = P • dS = i C —^ dSi, (10-183)
in agreement with Eq. (8-243) if C = —2i\. A negative sign is clearly
needed, since the viscous force opposes the velocity gradient. However,
Eq. (10-181) is not the most general linear relation between P and Vv
that is independent of the coordinate system. For we can further de-
compose (Vv), into a constant tensor and a traceless symmetric tensor
in the following way:
(Vv), = (Vv) c + (Vv)„, (10-184)
(Vv) c = i7V(W).l = iV • vl, (10-185)
(Vv)«. = (Vv), - iV • vl. (10-186)
This decomposition is independent of the coordinate system, since we
have shown that the trace is an invariant scalar quantity. We see by
Eq. (8-116) that the tensor (Vv)„ measures the rate of expansion or con-
traction of the fluid. The tensor (Vv) <8 , with five independent com-
ponents, specifies the way in which the fluid is being sheared. We are
therefore free to set
P = -2 v (Vv)» - WV • vl, (10-187)
with a coefficient r; which characterizes the viscous resistance to shear,
and a coefficient 17' which characterizes a viscous resistance, if any, to ex-
pansion and contraction. The last term corresponds to a uniform pressure
(or tension) in all directions at the given point. The coefficient i\' is small
and not very well determined experimentally for actual fluids. According
to the kinetic theory of gases, ij' is zero for an ideal gas. To the viscous
stress due to velocity gradients, given by formula (10-187), must be added
a hydrostatic pressure which may also be present and which depends on
the density, temperature, and composition of the fluid. If we lump the
last term in Eq. (10-187) together with the hydrostatic pressure into a
total pressure p, then the complete stress tensor is
P = pi - v [Vv + (Vv)' - §V • vl]. (10-188)
The reader may readily write this out in terms of components.
Formula (10-188) is the most general expression for the stress in an
isotropic fluid in which there is a hydrostatic pressure, plus viscous forces
proportional to the velocity gradient. It is possible to imagine that the
stress might also contain nonlinear terms in the velocity gradients, or even
high-order derivatives of the velocity, but such terms could be expected
444 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
to be small in comparison with the linear terms. Experimentally, the
viscous stresses in fluids are given very accurately in most cases by
formula (10-188).
We have decomposed the tensor Vv, with nine independent com-
ponents, into a sum of three tensors with one, three, and five independent
components, each a linear combination of the components of Vv. A
similar decomposition is clearly possible for any tensor. The reader may
well ask whether any further decomposition is possible. This is a problem
in group theory. We state without proof the result. Neither an anti-
symmetric tensor nor a symmetric traceless tensor can be further de-
composed in a manner independent of the coordinate system. The reader
can convince himself that this is plausible by a little experimentation.
Let us now consider an elastic solid. Let the solid be initially in an
unstrained position, and let each point in the solid be designated by its
position vector r relative to any convenient origin. Now let the solid be
strained by moving each point r to a new position given by the vector
r + p(r) relative to the same origin. We will designate the components
of r by (x, y, z) and of p by (£, j/, f). If p were independent of r, the mo-
tion would be a uniform displacement without deformation. Hence the
strain at any point may be specified by the gradient dyad
Vp
dx dx dx \
a| dv at
dy dy dy
\dz dz dz j
(10-189)
Now Vp can again be decomposed into an antisymmetric part which
corresponds to a rigid rotation about the point r + P and into a sym-
metric part which describes the deformation of the solid in the neighbor-
hood of each point :
S = iVp + i(Vp)'. (10-190)
The symmetric part can be further decomposed into a constant tensor
describing a volume compression or expansion and a symmetric traceless
tensor which describes the shear:
S C = JV- P 1 =i^Pl, (10-191)
S, t = £V P + i(Vp)' - Jv • pi. (10-192)
If the solid is isotropic, then this is the most general possible decomposi-
PROBLEMS 445
tion. Furthermore, if Hooke's law holds, the stress should be proportional
to the strain:
P = -|oV -pi —bS, t . (10-193)
The constants a and b are evidently related to the bulk modulus and the
shear modulus. If the solid is not isotropic, as for example, a crystal,
then the relation between P and S may depend on the choice of axes, and
must therefore be written:
Pa = Yj c iwSu. (10-194)
k,l=l
Since P and S have six independent components each, there are thirty-six
constants dju- By using the fact that there is an elastic potential energy
which is a function of the strain, it can be shown that in the most general
case there are twenty-one independent constants Cijki-
Problems
1. The product ct of a tensor by a scalar has been used in the text without
formal definition. Remedy this defect by supplying a suitable definition and
proving that this product has the expected algebraic properties.
2. Show that the centrifugal force in Eq. (7-37) is a linear function of the
position vector r of the particle, and find an expression for the corresponding
tensor in dyadic form. Write out the matrix of its coefficients.
3. Define time derivatives dt/dt and dTt/dt relative to fixed and rotating
coordinate systems, as was done in Chapter 7 for derivatives of vectors. Prove
that
where the cross product of a vector with a tensor is defined in the obvious way.
4. Write out the relations between the coefficients 0,7 corresponding to the
relations (10-80). Write down another relation between the unit vectors, in-
volving a triple cross product, and write out the corresponding relations be-
tween the coefficients.
5. Transform the tensor
T = AB -f- BA,
where
A = 5i - 3j + 2k, B = 5j + 10k,
into a coordinate system rotated 45° about the «-axis, using Eq. (10-74). Trans-
form the vectors A and B, using Eq. (10-71), and show that the results agree.
446 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
6. Write down and prove two additional relations like those in Eqs. (10-91)
through (10-93), involving algebraic properties which are preserved by trans-
formations of coordinates.
7. Prove Eqs. (10-91) and (10-92).
8. Write down the matrix for the orthogonal tensor A which produces a
rotation by an angle a about the 2-axis. Decompose A into a symmetric and
an antisymmetric tensor as in Eq. (10-66). What is the geometrical interpre-
tation of this decomposition?
9. Prove that Det (T) [Eq. (10-95)] is the same in all coordinate systems.
10. (a) Prove that the tensor given by formula (10-100) has the property
given by Eq. (10-23). (b) Prove by direct calculation that this tensor is repre-
sented by the same matrix in all coordinate systems.
11. Prove by direct calculation that the quantity M(T) denned by Eq. (10-139)
has the same value after the coordinate transformation (10-74).
12. Diagonalize the tensor in Problem 5. (That is, find its eigenvalues and
the corresponding principal axes.)
13. Diagonalize the tensor
(7 V6 -V3\
VQ 2 -5V2 J •
-\/3 — 5\/2 —3 /
(Hint: The secular equation can be factored; the roots are all integers.)
14. What are the principal axes and corresponding eigenvalues of the tensor
in Problem 2? Interpret physically.
15. Verify the statements made in the last paragraph of Section 10-4 regard-
ing the principal axis transformation found in the worked-out example.
16. Prove that if two tensors S and T have a set of principal axes in common,
then S • T = T • S. (The converse is also true.)
17. Prove that if a tensor T satisfies an algebraic equation
aj' n -\ h a 2 T 2 + aiT + a \ = 0,
where T"' means T • T • • • T (n factors), then its eigenvalues must satisfy the
same equation. The null tensor is defined in the obvious way.
18. Use the result of Problem 17 to show that the eigenvalues of the tensor
A representing a 180° rotation about some axis can only be ±1. [Hint: Consider
the result of applying A twice.] Show that the roots cannot all be +1. Then
show that —1 must be a double root. [Hint: Use Eqs. (10-137) and (10-81).]
Can you guess the corresponding eigenvectors? This entire problem is to be
answered by using general arguments, without writing down the matrix for A.
19. Show that the eigenvalues of an orthogonal tensor [Eq. (10-87)] are com-
plex (or real) numbers of unit magnitude. [Hint: Let C be an eigenvector (pos-
sibly complex) of T, and consider the quantity (T • C) • (T • C*).] Hence show
that one eigenvalue must be ±1, and the other two are of the form exp (±i a),
for some angle a.
PROBLEMS
447
20. Write out the components of the orthogonal tensor A corresponding to a
rotation by an angle 6 about the z-axis. Find its eigenvalues. Find and interpret
the eigenvectors corresponding to the real eigenvalue.
21. Find the components of the tensor corresponding to a rotation by an
angle 6 about the z-axis, followed by a rotation by an angle ip about the y-axis.
Find its eigenvalues. (Hint: According to Problem 19, one eigenvalue is ±1;
hence you can factor the secular equation.) Show that the result implies that
this transformation is equivalent to a simple rotation about some axis. (You
are not asked to find the axis.) Find the angle of rotation by comparing your
result with the eigenvalues found in Problem 20.
22. Show that the eigenvalues of an antisymmetric tensor are pure imaginary
(or zero) . Hence show that an antisymmetric tensor must have one zero eigen-
value and two conjugate imaginary eigenvalues. Find the eigenvectors corre-
sponding to the zero eigenvalue for the tensor (10-62).
23. Find the inertia tensor of a straight rod of length I, mass m, about its
center. Use this result to find the inertia tensor about the centroid of an equi-
lateral pyramid constructed out of six uniform rods. Show that this tensor can
be written down immediately from symmetry considerations, given the result
of Problem 17, Chapter 5.
24. Translate to the center of mass G the inertia tensor calculated about the
origin for the three disks in Fig. 10-2. Verify the statement made in the text
regarding the double degeneracy of lg.
25. Calculate the moment of inertia of a circular cone about a slant height.
[Hint: Calculate the inertia tensor about the apex relative to principal axes,
and use Eq. (10-143).]
26. Formulate and prove the most comprehensive theorem you can with
regard to the inertia tensor of a plane lamina. What can you say about the
principal axes and principal moments of inertia?
27. Find, by whatever method requires the least algebraic labor, the inertia
tensor of a uniform rectangular block of mass M , dimensions a X b X c, about
a set of axes through its center, of which the z-axis is parallel to side c, and the
2/-axis is a diagonal of the rectangle o X b.
28. (a) A uniform sphere of mass M , radius
a, has two point masses \M, ^M, located on
its surface and separated by an angular dis-
tance of 45°. Find the principal axes and
principal moments of inertia about the center
of the sphere, (b) Find the inertia tensor
about parallel axes through the center of
mass. Are they still principal axes?
29. (a) Find the inertia tensor of a plane
rectangle of mass M, dimensions a X 6. (b)
Use this result to find the inertia tensor about
the center of mass of the house of cards shown
in Fig. 10-5. Each card has mass M, dimen-
sions a X b(a < b). Use principal axes. Fig. 10-5. A house of cards.
448 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10
30. Find the equation for the ellipsoid of inertia of a uniform rectangular
block of dimensions I X w X h.
31. Find the equation for the ellipsoid of inertia of an object in the shape
of an ellipsoid whose equation is
2 2 2
I 2 ^ w 2 ^ h 2
32. Prove that the stress tensor P is symmetric.
33. Prove that if there is no shear on any surface element at some point, then
the stress tensor P at that point is a constant tensor [Eq. (10-23)].
34. Derive Eq. (10-171) by calculating the net force on a rectangular volume
element.
35. Derive from Eq. (10-173) the equation
^+V.(pw+P)=f,
which expresses the conservation of linear momentum. Show from this equation
that the momentum current tensor (pw + P) represents the flow of momentum,
and interpret physically the two terms in this tensor.
36. Derive from Eq. (10-172) a law of conservation of angular momentum
in a form analogous to Eq. (8-148).
37. Write the equations of motion (10-173) in cylindrical components for a
moving viscous fluid. Use these equations, together with suitable assumptions,
to derive Poiseuille's law (8-252) for steady viscous flow in a pipe. Write the
stress tensor for this case, in cylindrical coordinates, as a function of r, z, <p.
38. Write out the components of the stress tensor P in a viscous fluid.
39. (a) Show that the rate of production of kinetic energy per unit volume
due to stresses in a moving medium is
= -v(V-P).
(b) Show that the rate at which work is done by the stresses on the medium,
per unit volume, is
— =-V.(P-v).
{Hint: Calculate the work done across the surface of any volume V and use
Gauss' theorem.)
(c) Using these results, calculate the rate at which energy is dissipated per
unit volume by viscous stresses in a moving fluid. Write it out in terms of
components.
40. Find the relation between the constants a and b in Eq. (10-193) and the
bulk modulus B and shear modulus n defined by Eqs. (5-116) and (5-118).
[Set up S and P for the situations used in defining B and n.]
41. A solid is subject to a stress consisting of a pure tension t per unit area
in one direction. Find the strain S in terms of r, a, and 6. Using this, and the
PROBLEMS 449
result of Problem 40, express Young's modulus Y [Eq. (5-114)] in terms of B
and n. [Hint: Use symmetry to determine the form of S.]
*42. Find the most general linear relation between S and P for a nonisotropic
elastic substance which possesses cylindrical symmetry relative to a specified
direction.
*43. (a) Assume that in a nonisotropic elastic solid, there is an elastic potential
energy V per unit volume, which is a quadratic function of the strain com-
ponents. Show that there are 21 constants required to specify V.
(b) Show that if the strain in a solid in equilibrium is increased by SS, the work
done per unit volume against the stresses (exclusive of any work done against
body forces) is
$W J Pa dS fj :
[Hint: Calculate the work done on a volume element by the stresses on its
surface and use Gauss' theorem.]
(c) Combine results (a) and (b) to show that P is a linear function of S in-
volving 21 independent constants, in general.
cflP
dt :
= F,
dL
dt
= N,
P =
MV,
L =
l«,
CHAPTER 11
THE ROTATION OF A RIGID BODY
11-1 Motion of a rigid body in space. The motion of a rigid body in
space is determined by Eqs. (5-4) and (5-5) :
(11-1)
(H-2)
where
(11-3)
(11-4)
F and N are the total force on the body and the total torque about a
suitable point 0, V is the velocity of the center of mass, and I and w are
the inertia tensor and the angular velocity about the point 0. For an un-
constrained body moving in space, the point is to be taken as the center
of mass. If the body is constrained by external supports to rotate about
a fixed point, that point is to be taken as the point 0. If the point is
constrained to move in some fashion, the reader may supply the appro-
priate equation of motion. (See Chapter 7, Problem 3.)
Equations (11-2) and (11-4) for the rotation of a rigid body bear a
formal analogy to Eqs. (11-1) and (11-3) for the motion of a point mass
M. There are, however, three differences which spoil the analogy. In the
first place, Eq. (11-4) involves a tensor I, whereas Eq. (11-3) involves a
scalar M; thus P is always parallel to V, while L is not in general parallel
to u. A more serious difference is the fact that the inertia tensor I is not
constant with reference to axes fixed in space, but changes as the body
rotates, whereas M is constant (in Newtonian mechanics). Finally, and
perhaps most serious, is the fact that no symmetrical set of three coordi-
nates analogous to X, Y, Z exist with which to describe the orientation of
a body in space. This point was made in Section 5-1, and it is suggested
that the reader review the last paragraph in that section. For these rea-
sons, we cannot proceed to solve the problem of rotation of a rigid body
by analogy with the methods of Chapter 3.
There are two general approaches to the problem. We shall first, in
Sections 11-2 and 11-3, try to obtain as much information as possible
450
11-2] euler's equations of motion for a rigid body 451
from the vector equations (11-2), (11-4) without introducing a set of
coordinates to describe the orientation of the body. We shall then, in
Sections 11-4 and 11-5, use Lagrange's equations to determine the motion
in terms of a set of angular coordinates suggested by Euler.
11-2 Euler's equations of motion for a rigid body. The difficulty that
I changes as the body rotates may be avoided by referring Eq. (11-2) to
a set of axes fixed in the body. If we let "d'/dt" denote the time derivative
with reference to axes fixed in the body, then by Eq. (7-22), Eq. (11-2)
becomes
^ + «XL = N. (11-5)
Since I is constant relative to body axes, we may substitute from Eq. (11-4)
to obtain
I • — + « X (I • «) = N. (11-6)
(Recall that d'u/dt = du/dt.) It is most convenient to choose as body
axes the principal axes, ei, e 2 , e 3 of the body. Then Eq. (11-6) becomes
JlWl + (h — -J2)W 3 W2 = N U
1 2«2 + Vi — 7 3 )co 1 w 3 = N 2 , (11-7)
J3W3 + (I2 — Jl)<«>2«l = N 3 .
These are Euler's equations for the motion of a rigid body. If one point
in the body is held fixed, that point is to be taken as the origin for the body
axes, and the moments of inertia and torques are relative to that point.
If the body is unconstrained, the center of mass is to be taken as origin
for the body axes.
In order to derive the energy theorem from Euler's equations, we multi-
ply Eq. (11-6) by«:
u . 1 . 4» = „ . N. (11-8)
Since I is symmetric, the left member is
dw da . Id',.. dT . _,
B ' l 'I = I- | ' B= 2rf( (tt,|,B) = Hi' (U " 9)
where T is given by Eq. (10-153). Here we have used the fact that
d/dt, d'/dt have the same meaning when applied to a scalar quantity.
452 THE ROTATION OF A RIGID BODY [CHAP. 11
Comparing Eqs. (11-8) and (11-9), we obtain the energy theorem:
5 = « • N, (H-10)
at
in analogy with theorem (3-133) for the motion of a particle.
From Eqs. (11-7) we note immediately that a body cannot spin with
constant angular velocity w, except about a principal axis, unless external
torques are applied. If du/dt = 0, Eq. (11-6) becomes
to X (I • a) = N. (11-H)
The left member is zero only if I • <o is parallel to w, that is, if <o is along a
principal axis of the body. If a wheel is to spin freely without exerting
forces and torques on its bearings, then it must be not only statically
balanced, i.e., with its center of mass on the axis of rotation, but also
dynamically balanced, i.e., the axis of rotation must be a principal axis
of the inertia tensor, as any automobile mechanic knows.
In order to solve Eqs. (11-7) for u(t), we would need to know the com-
ponents of torque along the (rotating) principal axes, an uncommon situ-
ation, except for the case N = 0. We now consider a freely rotating sym-
metrical body, with no applied torque. Let the symmetry axis of the
body be e 3 , so that I\ = 7 2 - Then the third of Eqs. (11-7) is
/ 3 W3 = 0, (11-12)
and « 3 is constant. The first two equations may be written
"i + /3o>3« 2 = 0, <o 2 — |8«3«i = 0, (H-13)
where
18 = h 7 Zl • (H-14)
Equations (11-13) are a pair of coupled linear first-order equations in
o>i, <o 2 . Let us look for a solution by setting
»! = Axe 3 ", « a = A 2 e pt . (11-15)
We readily verify that formulas (11-15) satisfy Eqs. (11-13) provided that
p = ±tp» 8 , (11-16)
and
A 2 = Tiitj. (11-17)
11-2] euler's equations of motion for a rigid body 453
We have found a complex conjugate pair of solutions,
Wl = e ±tfi»,t f W2 = =F fe ±W-.« f (ii_i 8 )
and these may be superposed with arbitrary constant multipliers to form
the real solution :
ax = A cos (/3w 3 « + 0), co 2 = A sin (fio) 3 t + 8). (11-19)
The angular velocity vector a therefore precesses in a circle of radius A
about the e 3 -axis, with angular velocity /8w 3 . The precession is in the same
sense as o> 3 if I 3 > Ii, and in the opposite sense otherwise. The magni-
tude of <o is
co= [o>§ + A 2 ] 1 ' 2 , (H-20)
and is constant, a result which can also be proved by direct calculation of
d(co 2 )/dt from Eq. (11-7). The constants co 3 , A, 6 are determined by the
initial conditions. There are three arbitrary constants, since Euler's
equations are three first-order differential equations. Since an uncon-
strained rotating rigid body has three rotational degrees of freedom, we
should expect a total of six arbitrary constants to be determined by the
initial conditions. What are the other three?
The instantaneous axis of rotation, determined by the vector a, traces
out a cone in the body (the body cone) as it precesses around the axis of
symmetry. The half -angle a& of the body cone is given by
tana 6 =— • (11-21)
w 3
Alternatively, if the body is initially rotating with angular velocity «
about an axis making an angle aj, with the symmetry axis, then the con-
stants « 3 and A are given by
A = u sin ab, w 3 = w cos aj,. (11-22)
In order to find the motion in space, we need to locate the <o-axis with
respect to a direction fixed in space. We could do this by tracing out step
by step the motion relative to space axes, allowing the body to rotate
with constant angular velocity about an axis in the body cone which
precesses with angular velocity /3to 3 . It is easier to locate w relative to
L, since by Eq. (11-2) L is constant if N = 0. The angle a„ between <o
and L is given by
«-L <ol« 2T . .
cos a > = -^l = —^r = ^l • (11_23)
Since by Eq. (11-10) T is constant, the angle a s is constant. The axis of
454 THE ROTATION OF A RIGID BODY [CHAP. 11
rotation therefore traces out a cone in space, the space cone. The space
cone has a half-angle a a given by Eq. (11-23) and its axis is the direction
of the angular momentum vector L. The line of contact between the
space cone and the body cone at any instant is the instantaneous axis of
rotation. Since this axis in the body is instantaneously at rest, the body
cone rolls without slipping around the space cone. This gives a complete
description of the motion (see Fig. 11-1).
We can express a a in terms of the constants co, a&. We have
I = (eiei + e 2 e 2 )/i + e 3 e 3 / 3
= hi + e 3 e 3 (/ 3 - h). (H-24)
By substituting from Eqs. (11-14), (11-22), and (11-24) into Eqs. (11-4),
(10-153), and (11-23), and with u = wn, we obtain:
2T = w 2 /i[l + p cos 2 a b ], (11-25)
L = ali[n + P cos a 6 e 3 ], (11-26)
1 + ft cos 2 a b n , _,
C0S "> = [1 + (20 + ]8»)cos» « 6 ]i/» ' (11_27)
Note that a„ depends on only a& and not on w. It is clear from Eq. (11-26)
that the space cone lies inside the body cone if > and outside if /3 <
(see Fig. 11-1). This is clear also if, when the body cone rolls on the space
cone, the precession of the axis of rotation is to have the sense given by
Eq. (11-19). [The reader should check this, remembering that Eq. (11-19)
describes the motion of the axis relative to the body.]
Next, consider the case when the inertia tensor is nondegenerate. We
shall number the principal axes so that 7 3 > 7 2 > Ii- It was shown
above that a body may rotate freely about a principal axis. Let us study
small deviations from this steady rotation. If <o is not along a principal
axis, then it cannot remain constant. Let us assume that u lies very close
to a principal axis, say to e 3 , so that o> 3 ~2> o>i and w 3 » o> 2 . Then if
N = 0, we see from the third of Eqs. (11-7) that w 3 is constant to first
order in «i and w 2 . The first two equations then become a pair of coupled
linear equations in u\, co 2 , which we solve as in the preceding example
to obtain
«i = A[I 2 (I 3 - / 2 )] 1/2 cos (/3w 3 < + 0),
o> 2 = Alhih - h)] 1 ' 2 sin ((3a> 3 t + B),
(11-28)
where A and 6 are arbitrary constants and
p = [ (J3 ~ Z %\ ~ /2) ] 1/2 - dl-29)
11-3]
poinsot's solution for a freely rotating body
455
Fig. 11-1. Free rotation of a symmetrical body.
The vector w therefore moves counterclockwise (looking down from the
positive e 3 -axis) in a small ellipse about the e3-axis. In a similar manner,
we can show that if u is nearly parallel to the ei-axis, it moves clockwise
in a small ellipse about that axis, and that if a is nearly parallel to the
e2-axis, the solution is of an exponential character. In the latter case, of
course, the components «i and a>3 will not remain small, and the approxi-
mation that «2 is constant will hold only during the initial part of the
motion. We conclude that rotation about the axes of maximum and
minimum moments of inertia is stable, while rotation about the inter-
mediate axis is unstable. This result is readily demonstrated by tossing
a tennis racket in the air and attempting to make it spin about any prin-
cipal axis. The general solution of Eqs. (11-7) for u, when N = 0, can
also in principle be obtained. We shall solve the problem in the next
section by a different method.
11-3 Poinsot's solution for a freely rotating body. If there are no
torques, N = 0, then Eqs. (11-2) and (11-10) yield four integrals of the
equations of motion:
L = I • w = a constant, (11-30)
T = \ <o • I • «
a constant.
(11-31)
456 THE ROTATION OF A RIGID BODY [CHAP. 11
Poinsot* has obtained a geometrical representation of the motion based
on these constants, and utilizing the inertia ellipsoid. Let us imagine the
inertia ellipsoid (10-157) rigidly fastened to the body and rotating with
it. If we let r be the vector from the origin to the point where the axis of
rotation intersects the inertia ellipsoid at any instant,
r
- «,
CO
then comparison of Eqs. (11-31) and (10-157) shows that
2 2
a co
(11-32)
(11-33)
2r 2
The normal to the ellipsoid at the point r is parallel to the vector
V(r • I • r) = 2I i x 1 e 1 + 2I 2 x 2 e 2 + 2I 3 x 3 e 3 = 2 £ L, (11-34)
where x x , x 2 , x 3 are the components of r along the principal axes. The
tangent plane to the ellipsoid at the point r is therefore perpendicular to
the constant vector L (see Fig. 11-2). Let I be the perpendicular distance
from the origin to this tangent plane:
r-L r co-l-w a(2T) 112 . , , 1t „-
I = -—=- = = = v T ' — = a constant. (11-35)
The tangent plane is therefore fixed in space (relative to the origin 0) and
is called the invariable plane. Its position is determined by the initial con-
ditions. Moreover, since the point of contact between the ellipsoid and
the plane lies on the instantaneous axis of rotation, the ellipsoid rolls on
the plane without slipping. The angular velocity at any instant has the
magnitude
(2r)1/2 r. (11-36)
co =
This gives a complete description of the motion. ^
As the inertia ellipsoid rolls on the invariable plane, with its center
fixed at the origin, the point of contact traces out a curve called the
polhode on the inertia ellipsoid, and a curve called the herpolhode on the
invariable plane. This is illustrated in Fig. 11-2. The polhode is a closed
curve on the inertia ellipsoid, denned as the locus of points r where the
tangent planes lie a fixed distance I from the center of the ellipsoid. In
Fig. 11-3 are shown various polhodes on a nondegenerate inertia ellipsoid.
Note that the topological features of the diagram are in agreement with
* Poinsot, Theorie Nouvelle de la Rotation des Corps, 1834.
11-3] poinsot's solution for a freely rotating body 457
Inertia ellipsoid
Herpolhode /Invariable plane
Fig. 11-2. The inertia ellipsoid rolls on the invariable plane.
Fig. 11-3. Polhodes on a nondegenerate inertia ellipsoid.
the conclusions at the end of the preceding section. In general, the herpol-
hode is not closed but fills an annular ring in the invariable plane.
In the case of a symmetrical body it can be shown (Problem 7) that the
polhodes are circles about the symmetry axis and the herpolhodes are
circles in the invariable plane. In that case, r and therefore, by Eq. (11-36),
co (but not to!) are constant during the motion. Poinsot's description of
the motion in this case agrees with that in the preceding section. The
458
THE ROTATION OF A RIGID BODY
[CHAP. 11
polhode and herpolhode are the intersections of the body and the space
cones with the inertia ellipsoid and the invariable plane, respectively.
11-4 Euler's angles. The results in Sections 11-2 and 11-3 regarding
the motion of a rigid body were obtained without the use of any coordi-
nates to describe the orientation of the body. In order to proceed further
with the discussion, it is necessary to introduce a suitable set of coordi-
nates. We choose a set of axes fixed in the body, which are most con-
veniently taken as the principal axes, with origin at the center of mass,
or at the fixed point if one exists. These axes will be labeled with sub-
scripts 1, 2, 3, as before. If there is an axis of symmetry, it will be num-
bered 3; otherwise, the axes may be numbered in any order. We need
three coordinates to specify the orientation of the body axes with respect
to a fixed set of space axes x, y, z. The relation between the two sets of
axes could be specified by giving the coefficients of the transformation
from coordinates x, y, z to xi, x 2 , x 3 . There are nine coefficients but only
three of them are independent, as we have seen, and to try to use three of
the coefficients as coordinates is not convenient. As was pointed out in
Section 5-1, there is no symmetric set of coordinates analogous to x, y, z
with which to describe the orientation of a body. Among the various
coordinate systems that have been introduced for this purpose, one of
the most useful is due to Euler.
In Fig. 11-4, the Euler angles 6, <j>, $ are shown. These are used to spec-
ify the position of the body axes 1, 2, 3 relative to the space axes x, y, z.
The body axes 1, 2, 3 are shown as heavy lines; the space axes x, y, z are
lighter. The angle is the angle between the 3-axis and the z-axis.
Since the 3-axis is thus singled out for special treatment, if the body has
an axis of symmetry, it should be taken as the 3-axis. Likewise, if the
external torques possess an axis of symmetry in space, that axis should
be taken as the z-axis. The intersection of the 1, 2-plane with the x?/-plane
Fig. 11-4. Euler's angles.
11-4] euler's angles 459
is called the line of nodes, labeled £ in the diagram. The angle <f> is meas-
ured in the xy-pla,ne from the z-axis to the line of nodes, as shown. The
angle ^ is measured in the 1, 2-plane from the line of nodes to the 1-axis.
We are assuming that both sets of axes x, y, z and 1, 2, 3 are right-handed.
It will be convenient also to introduce a third (right-handed) set of axes,
£, ri, f, of which £ is the line of nodes, f coincides with the body axis 3,
and ij is in the 1, 2-plane. |
In order to express the angular velocity vector o> in terms of Euler's
angles, we first prove that angular velocities may be added like vec-
tors, in the sense of the following theorem:
Given a primed coordinate system rotating with angular velocity
«i with, respect to an unprimed system, and a starred coordinate
system rotating with angular velocity u 2 relative to the primed
system, the angular velocity of the starred system relative to the
unprimed system is «i + «2. (11-37)
To prove this theorem, let A be any vector at rest in the starred system:
Then by theorem (7-22), its velocity relative to the primed system is
"=« 2 XA. (11-39)
Now applying theorem (7-22) again, we find the velocity of A relative to
the unprimed system:
^=^+ Wl XA=(<o 1 + » 2 )XA. (llHtO)
A final comparison with theorem (7-22) shows that («i + «2) is the
angular velocity of the starred system relative to the unprimed one.
Now consider Figure 11-4 and suppose that the body is moving so that
6, <j>, ^ are changing with time. If alone changes, while <£, \p are fixed,
the body rotates around the line of nodes with angular velocity tej. If <j>
alone changes, the body rotates around the z-axis with angular velocity
<j>k. If ^ alone changes, the body rotates around its 3-axis with angular
velocity ^e 3 . Now if we consider a primed coordinate system rotating with
f The reader is cautioned that the notation for Euler's angles, as well as the
convention as to axes from which they are measured, and even the use of right-
handed coordinate axes, are not standardized in the literature. It is therefore
necessary to note carefully how each author defines the angles. The conventions
adopted here are very common, but not universal.
460 THE ROTATION OF A RIGID BODY [CHAP. 11
angular velocity <£k about the 2-axis, and let the i-,»j,f-system rotate with
angular velocity 0ej relative to this primed system, then by theorem
(11-37), the angular velocity of the £,ij,f-system is 0e { + <£k. The axes
1, 2, 3 rotate with angular velocity ^e 3 relative to £,ij,f, hence the angular
velocity of the body is
« = fe{ + <£k + ^e 3 . (11-41)
We have, from Fig. 11-4, the relations
ej = ei cos ^ — e 2 sin ^,
e, = ei sin ^ + e 2 cos if/, (11-42)
e r = e 3 ,
and
k = e ? cos 6 + e, sin 6
= ei sin 6 sin \p + e 2 sin B cos 4> + e 3 cos 6. (11-43)
We may therefore express a in terms of its components along the principal
axes:
Wi = 6 cos ^ + <j> sin 6 sin \[/,
w 2 = —6 sin $ + <j> sin cos ^, (11-44)
w 3 = ^ + 4> cos 0.
The kinetic energy is now given by Eq. (10-153) :
T = i/i«? + \U<4 + */*»§. (11-45)
The kinetic energy is a rather complicated expression involving 6, <j>, 4>, 0,
and 4/. Note that 0, <£, ^ are not orthogonal coordinates, i.e., cross terms
involving 6<j> and ^ appear in T. In the case of a symmetrical body
(Ii = 7 2 ), the expression for T simplifies to the form:
T = %I 1 6 2 + i/^ 2 sin 2 + J/gGJ- + <£ cos 0) 2 . (11-46)
The generalized forces Qe, Q«, Q* are easily shown to be the torques
about the £-, z-, and 3-axes.
We are now in a position to write down Lagrange's equations for the
rotation of a rigid body subject to given torques. If the torques are de-
rivable from a potential energy V(6,<j>,ip), then there will be an energy
integral. If V is independent of <f>, then inspection of Eqs. (11-44) shows
that <j> will be an ignorable coordinate. Unfortunately, this is not enough
to enable us to give a general solution of the problem. However, for a
symmetrical body, if V is independent of yp also, we see from Eq. (11-46)
11-5}
THE SYMMETRICAL TOP
461
that both <f> and $ are ignorable. We have then three constants of the
motion, enough to solve the problem. This case will be solved in the
next section. A few other special cases are known for which the problem
can be solved,* but for the general problem of the motion of an unsym-
metrical body under the action of external torques, as for the many-body
problem, there are no generally applicable methods of solution, except by
numerical integration of the equations of motion.
11-5 The symmetrical top. The symmetrical top, represented in
Fig. 11-5, is a body for which Ii = 7 2 . It pivots around a fixed point
that lies on the axis of symmetry a distance I from the center of mass G
which also lies on the axis of symmetry. The only external forces are the
forces of constraint at and the force of gravity. Therefore, by Eq. (11-46),
the Lagrangian function is
L = \I X P + J/^ 2 sin 2 6 + J/ 8 (tf + cos 0) 2 - mgl cos 8. (11-47)
The coordinates ^ and <j> are ignorable, and we have therefore three
integrals of the motion:
dpj, dL
~dt ~ w
0,
where
d£* = — = n
dt d<t>
dE__dL =a
dt ~~ dt ~ '
V* = ^(^ + 4> cos 0),
p* = Ii4> sin 2 + 7 3 cos 6(4/ + 4 cos 9),
E = ilj 2 + J/^ 2 sin 2 6
+ fclaOA + <£ cos 0) 2 + mgl cos 0.
(11-48)
(11^9)
(11-50)
(11-51)
(11-52)
(11-53)
We use Eqs. (11-51) and (11-52) to eliminate i, <£ from Eq. (11-53):
E = w2 + (?*-^ n y) 2 + A + ^ cos ,. (11 -54)
* See, for example, E. J. Routh, The Advanced Part of a Treatise on the Dy-
namics of a System of Rigid Bodies, 6th ed. London: Macmillan, 1905. (Also
New York: Dover, 1955.)
462
THE ROTATION OF A RIGID BODY
[CHAP. 11
Fig. 11-5. Coordinates for the symmetrical top.
We can now solve the problem by the energy method. If we set
W = E -
~2
V*
2J»
(P* ~ V* cos 0) ''
+ mgl cos 6,
then
2/i sin* 6
= \j-[E' -<F(f»]} 1/2 ,
and 6 is given, in principle, by computing the integral
f * =(L\» t
Je [E' - T(8)]W \2y/
(11-55)
(11-56)
(11-57)
(11-58)
and solving for 6(t). The constant O is the initial value of 0. Once 6{t)
is known, Eqs. (11-51) and (11-52) can be solved for ^ and <f> and inte-
grated to give ^(t), <t>(t).
Comparison of Eqs. (11-44) and (11-51) shows that
p+ = I3013,
(11-59)
so that w 3 is a constant of the motion. If W3 = 0, then Eq. (11-56) re-
duces essentially to the formula (9-137) for a spherical pendulum, as it
11-5]
THE SYMMETRICAL TOP
463
0o */2
Fig. 11-6. Effective potential energy for the symmetrical top.
should. In Fig. 11-6, 'F'(0) is plotted versus for o> 3 ^ 0. The 'torque'
associated with the 'potential energy' 'V'(d) is
•N' = -
d'V
mgl sin —
(p» — V* cos 0)(p» — p» cos 0)
1 1 sin3
(11-60)
Inspection of Eq. (11-60) shows that, in general (if p* j£ p+), the 'torque'
W is positive f or 8 = and negative for = 7r, and has one zero between
and ir. Hence 'V has one minimum, as shown in Fig. 11-6, at a point 6
satisfying the equation
mgl 1 1 sin 4 O — (p# — V* cos o )(p* — P* cos O ) = 0. (11-61)
If E' = 'V'(8 ), the axis of the top precesses uniformly at an angle O
with the vertical, and with angular velocity
P* ~ V* cos flp
1 1 sin 2 O
<£o
(11-62)
Solving Eq. (11-61) for (p* — p* cos O ), and using Eq. (11-59), we
obtain
(p — p* COS O ) = i/ 3 W3
sin 2 0p
cos 0Q
(11-63)
We see that if 0o < n"/2, there is a minimum spin angular velocity below
which the top cannot precess uniformly at the angle O :
Wmin = J ^ COS O J
(11-64)
464 THE ROTATION OF A RIGID BODY [CHAP. 11
For « 3 > w min , there are two roots (11-63) and hence two possible values
of <>o, a slow and a fast precession, both in the same direction as the spin
angular velocity « 3 . For w 3 5>> w m i n , the fast and slow precessions occur
at angular velocities
4, ^h J**- (11-65)
vo 7i cos O
4,0 ~j£l' (ll_66)
and
It is the slow precession which is ordinarily observed with a rapidly spin-
ning top. For O > t/2 (top hanging with its axis below the horizontal),
there is one positive and one negative value for 4> . (To what do these
motions of uniform precession reduce when oo s — » 0?)
Study of Fig. 11-6 shows us that the more general motion involves a
nidation or oscillation of the axis of the top in the 0-direction as it pre-
cesses. The axis oscillates between angles 0i and 2 which satisfy the
equation
E> = ^-jX^ + m91 C0S *> {U ^ 7)
where p^,, p+, and E' are determined from the initial conditions. If we
multiply Eq. (11-67) by sin 2 0, it becomes a cubic equation in cos 0. We
see from Fig. 11-6 that there must be two real roots cos 1; cos 2 between
—1 and +1. The third root for cos must lie outside the physical range
— 1 to +1. In fact, inspection of Eq. (11-67) will show that the third
root is greater than +1. (In the case of uniform precession discussed
in the preceding paragraph, the two physical roots coincide, cos 0i =
cos 2 = cos O .) If initially 0=0, then the initial value cos 0i of cos
satisfies Eq. (11-67) ; knowing one root of a cubic equation, we may factor
the equation and find all three roots. During nutation, the precession
velocity varies according to Eq. (1 1-52) :
1 _ P* — P+ COS fii_RO\
+ - hsm'0 ' (11_68)
If \p+\ < \p+\, we can define an angle 3 as follows:
cos 3 = £* • (11-69)
For > 3 , 4> has the same sign as w 3 , and for < 3 , it has an opposite
sign. The derivative with respect to of the right member of Eq. (11-67)
is negative at = 3 ; hence we see from Fig. 11-6 that 3 < 02, where
2 is the largest angle satisfying Eq. (11-67). In fact, 3 < O . If 3 < 0i
11-5]
THE SYMMETRICAL TOP
465
(a) (b) (c)
Fig. 11-7. Locus of top axis (3) on unit sphere.
(or if |p^| > |p*| and p^, p* have the same sign), then <j> has the same sign
as « 3 throughout the nutation, and the top axis traces out a curve like
that shown in Fig. 11-7 (a). If 6 3 > U <£ changes sign during the nutation
and the top axis moves as in Fig. ll-7(b). It is clear that if the top is set
in motion initially above the horizontal plane with <f> opposite in sign to
« 3 , the motion necessarily will be like that shown in Fig. ll-7(b).
An important special case occurs when the top, spinning about its axis
with angular velocity w 3 , is held with its axis initially at rest at an angle
0i and then released. Initially, we have
e = e u 6 = o, 4> = o, ^ = w 3 . (n-70)
We substitute in Eqs. (11-51), (11-52), and (11-53) to find
pj, = 7 3 W3, p* =. I 3 u 3 cos 0i, E' = mgl cos Oi. (11-71)
In this case, we see that 8 3 = B\, and the motion is as shown in Fig. 11-7 (c).
An elementary discussion of this case, based on the conservation of angu-
lar momentum, was given in Section 4-2. Now Eq. (11-56) becomes
'V
where
2/i L
(cos 6 1 — cos fl) 2
sin 2 6
2Iimgl
i%<4 '
+ a cos 6
]■
(11-72)
(11-73)
The turning points for the nutation are the roots of Eq. (11-67), which
becomes in this case, if we multiply by sin 2 6,
(cos 8 1 — cos 6) 2 — a(cos 8 y — cos 0)(1 — cos 2 6) = 0. (11-74)
466 THE ROTATION OF A RIGID BODY [CHAP. 11
The roots are
cos 6 = cos $i,
(11-75)
cos = ± [1 ± (1 - 4a cos 0i + 4a 2 ) 1 ' 2 ].
2a
The angle 2 is given by the second formula, using the minus sign in the
bracketed expression. The plus sign gives a root for cos greater than
+1. Let us consider the case of a rapidly spinning top, that is, when
a « 1. We then have
cos 02 = cos 0i — a sin 2 X . (11-76)
The angle 2 is only slightly greater than 0i, and the amplitude of nuta-
tion is proportional to a. If we set
2 = 0o + a, 0i = O — a, (11-77)
and substitute in Eq. (11-76), we find that, to first order in a and a,
a = \a sin 0i. (11-78)
We now set
6 = O + S = 0i + a + 8, (11-79)
and substitute in Eq. (11-72), which becomes, to second order in a and 5,
t2
'V = F(0 O ) + i ^ «! S 2 . (11-80)
^i
The first term is constant, and the second leads to harmonic oscillations
in 3 with a frequency
<o 9 = ^ « 3 . (H-81)
The nutation is given by
= 0i + a — acosaet. (11-82)
We substitute in Eq. (11-68) to obtain <j> to first order in a:
1 1 sin Pi
The average angular velocity of precession is
W „ * ^tfa« "KL (11 _ 84 )
XYVa Ii sin 0i /3W3
11-5]
THE SYMMETRICAL TOP
467
Fig. 11-8. Effective potential energy when p^ = p+.
The top axis therefore precesses very slowly and nutates very rapidly with
very small amplitude. In practice, the frictional torques which we have
neglected usually damp out the nutation fairly quickly, leaving only the
uniform precession.
As a final example, consider the case when the top is initially spinning
with its symmetry axis vertical. In this case, so long as the 3- and z-axes
coincide, the line of nodes is indeterminate. We see from Fig. 11-4 that
the angle ^ + <f> is determined as the angle between the x- and 1-axes,
although \f/, <t> separately are indeterminate. Hence we have initially
V* = liU + 4) = -*3W3,
P+ = l3(t + 4>) = V+-
(11-85)
(11-86)
Equation (11-56) in this case becomes
, v , _ jj»ira
¥ ~~ 27i L tin* 9
cos 0)
+ a cos 6
(11-87)
where a is given by Eq. (11-73). This, of course, is just a special case of
Eq. (11-72). In Fig. 11-8 we plot 'V for the case when p* = p*. The
form of the curve depends upon the value of a. We see that a rapidly
spinning top (a < £) can spin stably about the vertical axis; if disturbed,
it will exhibit a small nutation about the vertical axis. A slowly spinning
top (a > J) cannot spin stably about a vertical axis, but will execute a
large nutation between 0i = and 8 2 given by Eq. (11-75). In this case,
cos 6 2 =
(11-88)
468 THE ROTATION OF A RIGID BODY [CHAP. 11
The minimum spin angular velocity below which the top cannot spin
stably about a vertical axis occurs when a = \, or, by Eq. (11-73),
Wmin - hr.
(11-89)
Note that this formula agrees with Eq. (11-64). If initially w 3 > w m i n ,
a top will spin with its axis vertical, but when friction reduces w 3 below
w m in, it will begin to wobble.
All of the above conclusions about the behavior of a symmetrical top
under various initial conditions can easily be verified experimentally with
a top or with a gyroscope.
Problems
1. Use the result of Problem 3, Chapter 10, to derive Eq. (11-6) directly from
the equation
2. (a) Assume that the earth is a uniform rigid ellipsoid of revolution, look
up its equatorial and polar diameters, and calculate the angular velocity of
precession of the North Pole on the earth's surface assuming that the polar axis
(i.e., the axis of rotation) deviates slightly from the axis of symmetry. (An
irregular precession of roughly this sort is observed with an amplitude of a few
feet, and a period of 427 days.)
(b) Assume that the earth is a rigid sphere and that a mountain of mass 10~ 9
times the mass of the earth is added at a point 45° from the polar axis. Describe
the resulting motion of the pole. How long does the pole take to move 1000
miles?
(c) For a rigid ellipsoidal earth, as in part (a), how massive a "mountain"
must be placed on the equator in order to make the polar precession unstable?
The earth is, of course, not of uniform density, but is more dense near its
center. Even more important, the earth is not rigid, but behaves as an elastic
spheroid for short times, and can deform plastically over long times. The results
in this problem are therefore only suggestive and do not correspond to the
actual motion of the earth. For example, the observed precession period of 427
days is longer than would be calculated for a rigid earth. When plastic deforma-
tion is taken into account, an appreciable wandering of the pole can result
even for an ellipsoidal earth with a much smaller "mountain" than that cal-
culated in part (c).*
* An excellent short discussion of the rotation of the earth, treated as an
elastic and plastic ellipsoid, will be found in an article by D. R. Inglis, Review
of Modern Physics, vol. 29, p. 9 (1957).
PROBLEMS 469
3. Show that the axis of rotation of a freely rotating symmetrical rigid body
precesses in space with an angular velocity
«n = (|3 + sec a&)co3,
where the notation is that used in Section 11-2.
4. Show that if the only torque on a symmetrical rigid body is about the axis
of symmetry, then (<u? + o>!) is constant, where «i and a>2 are angular velocity
components along axes perpendicular to the symmetry axis. If #3(0 is given,
show how to solve for coi, 012, and C03.
5. A symmetrical rigid body moving freely in space is powered with jet engines
symmetrically placed with respect to the 3-axis of the body, which supply a
constant torque JV3 about the symmetry axis. Find the general solution for the
angular velocity vector as a function of time, relative to body axes, and describe
how the angular velocity vector moves relative to the body.
6. (a) Consider a charged sphere whose mass m and charge e are both dis-
tributed in a spherically symmetrical way. Show that if this body rotates in a
uniform magnetic field B, the torque on it is
N = rr^- L X B (gaussian units),
2mc
where g is a numerical constant, which is one if the mass density is everywhere
proportional to the charge density.
(b) Write an equation of motion for the body, and show that by introducing
a suitably rotating coordinate system, you can eliminate the magnetic torque.
(c) Compare this result with Larmor's theorem (Chapter 7). Why is no
assumption needed here regarding the strength of the magnetic field?
(d) Describe the motion. What points in the body are at rest in the rotating
coordinate system?
7. Prove (without using the results of Section 1-2) that if two principal
moments of inertia are equal, the polhode and the herpolhode are both circles.
8. (a) Obtain equations, in terms of principal coordinates xi, X2, xz, for two
quadric surfaces whose intersection is the polhode. Your equations should
contain the parameters 7i, I2, I3, I-
(b) Find the equation for the projection of the polhode on any coordinate
plane and show that the polhodes are closed curves around the major and minor
poles of the ellipsoid, but that they are of hyperbolic type near the intermediate
axis, as shown in Fig. 11-3.
(c) Find the radii of the circles on the invariable plane which bound the
herpolhode.
9. Find the matrix (ay) which transforms the components of a vector from
space axes to body axes. Express an in terms of Euler's angles. [Hint: The
transformation can be made up of three consecutive rotations by angles 9, 4>> lA>
about suitable axes, and taken in proper order.]
10. Write out the Hamiltonian function in terms of 6, \p, <£, p«, p*, p* for a
freely rotating unsymmetrical rigid body. Express the coefficients in terms
of the parameters Ii, I3, (I2 — li)-
470 THE BOTATION OF A RIGID BODY [CHAP. 11
11. Use Lagrange's equations to treat the free rotation of an unsymmetrical
rigid body near one of its principal axes, and show that your results agree with
the last paragraph of Section 11-2.
12. Set up Lagrange's equations for a symmetrical top, the end of whose
axis slides without friction on a smooth table. Discuss carefully the differences
in the motions between this case and the case when the end of the top axis pivots
about a fixed point.
13. A gyroscope is constructed of a disk of radius a, mass M, fastened rigidly
at the center of an axle of length (3a/2), mass (2M/7), negligible cross section,
and mounted inside two perpendicular rings, each of radius (3o/2), mass (M/3).
The axle rotates in frictionless bearings at the intersection points of the rings.
One of these intersection points pivots without friction about a fixed point 0.
Set up the Lagrangian function and discuss the kinds of motion which may
occur (under the action of gravity) .
14. Discuss the free rotation of a symmetrical rigid body, using the La-
grangian method. Find the angular velocity for uniform precession and the
frequency of small nutations about this uniform precession. Describe the mo-
tion and show that your results agree with the solutions found in Section 11-2
and in Problem 3.
15. A top consists of a disk of mass M, radius r, mounted at the center of a
cylindrical axle of length I, radius a, where a <3C I, and negligible mass. The end
of the axle rests on a table, as shown in Fig. 11-9. The coefficient of friction
is fi. The top is set spinning about its symmetry axis with a very great angular
velocity «3o, and released with its axis at an angle 0i with the vertical. Assume
that C03 is great enough compared with all other motions of the top so that the
edge of the axle in contact with the table slides on the table in a direction per-
pendicular to the top axis, with the sense determined by «3. Write the equa-
tions of motion for the top. Assume that the nutation is small enough to be
neglected, and that the friction is not too great, so that the top precesses slowly
at an angle 0o which changes slowly due to the friction with the table. Show
Fig. 11-9. A simple top.
PROBLEMS 471
that the top axis will at first rise to a vertical position, and find approximately
the time required and the number of complete revolutions of precession during
this time. Describe the entire motion of the top relative to the table during
this process. How long will it remain vertical before beginning to wobble?
16. Obtain a toy gyroscope, and make the necessary measurements in order to
predict the rate at which it will precess, when spinning at its top speed, if its
axis pivots about a fixed point at an angle of 45° with the vertical. Calculate
the amplitude of nutation if the axis is held at an angle of 45° and released. Per-
form the experiment, and compare the measured rate of precession with the
predicted rate.
17. A planet consists of a uniform sphere of radius a, mass M, girdled at its
equator by a ring of mass m. The planet moves (in a plane) about a star of
mass M' . Set up the Lagrangian function, using as coordinates the polar co-
ordinates r, a in the plane of the orbit, and Euler's angles 6, <t>, yf/, relative to
space axes of which the z-axis is perpendicular to the plane of the orbit, and the
x-axis is parallel to the axis from which a is measured. You may assume that
r ^> a, and use the result of Problem 13, Chapter 6. Find the ignorable co-
ordinates, and show that the period of rotation of the planet is constant.
18. Assume that the planet of Problem 17 revolves in a circle of radius r about
the star, although this does not quite satisfy the equations of motion. Assume
that the period of revolution is short in comparison with any precession of the
axis of rotation, so that in studying the rotation it is permissible to average over
the angle a. Show that uniform (slow) precession of the polar axis may occur if
the axis is tilted at an angle 0o from the normal to the orbital plane, and find
the angular velocity of precession in terms of the masses M, m, M', the radii
a, r, the angle 0o, and the angular velocity of rotation. Show that if the day is
much shorter than the year, the above assumption regarding the period of revo-
lution and the rate of precession is valid. Find the frequency of small nutations
about this uniform precession and show that when the day is much shorter
than the year, it corresponds to the free precession whose angular velocity is
given in Problem 3.
19. Find the masses M, m required to give the planet in Problem 17 the same
principal moments of inertia as a uniform ellipsoid of the same mass and shape
as the earth. Show that, with the approximations made in Problem 18, if the
sun and moon lie in the earth's orbital plane (they do very nearly), the effect
of both sun and moon on the earth's rotation can be taken into account simply
by adding the precession angular velocities that would be caused by each
separately. The equator makes an angle of 23.5° with the orbital plane. Find
the resulting total period of precession. (The measured value is 26,000 years.)
*20. Write Lagrangian equations of motion for the rigid body in Problem 5.
Carry the solution as far as you can. (Make use of the results of Problem 5 if
you wish.) Show that you can obtain a second order differential equation in-
volving d alone. Can you find any particular solutions, or approximate solutions,
of this equation for special cases? Describe the corresponding motions. (Note
that this problem, to the extent that it can be solved, gives the motion of the
body in space, in contrast to Problem 5, where we found the angular velocity
relative to the body.)
472 THE ROTATION OF A RIGID BODY [CHAP. 11
21. An electron may for some purposes be regarded as a spinning charged
sphere like that considered in Problem 6, with g very nearly equal to 2. Show
that if g were exactly 2, and the electron spin angular momentum is initially
parallel to its linear velocity, then as the electron moves through any magnetic
field, its spin angular momentum would always remain parallel to its velocity.
22. An earth satellite consists of a spherical shell of mass 20 kgm, diameter
1 m. It is directionally stabilized by a gyro consisting of a 4-kgm disk, 20 cm
in diameter, mounted on an axle of negligible mass whose frictionless bearings
are fastened at the opposite ends of a diameter of the shell. The shell is initially
not rotating, while the gyro rotates at angular velocity wo- A one-milligram
dust grain traveling perpendicular to the gyro axis with a velocity of
3 X 10 4 m/sec buries itself in the shell at one end of the axis. What must be
the rotation frequency of the gyro in order that the gyro axis shall thereafter
remain within 0.1 degree of its initial position? An accuracy of two significant
figures in the result will be satisfactory.
23. A gyrocompass is a symmetrical rigid body mounted so that its axis is
constrained to move in a horizontal plane at the earth's surface. Choose a
suitable pair of coordinate angles and set up the Lagrangian function if the
gyrocompass is at a fixed point on the earth's surface of colatitude 0o- Neglect
friction. Show that the angular velocity component &>3 along the symmetry
axis remains constant, and that if «3 > (Ii/hfao sin do, where wo is the angular
velocity of the earth, then the symmetry axis oscillates in the horizontal plane
about a north-south axis. Find the frequency of small oscillations. In an actual
gyrocompass, the rotor must be driven to make up for frictional torques about
the symmetry axis, while frictional torques in the horizontal plane damp the
oscillations of the symmetry axis, which comes to rest in a north-south line.
CHAPTER 12
THEORY OF SMALL VIBRATIONS
An important and frequently recurring problem is to determine whether
a given motion of a dynamical system is stable, and if it is, to determine
the character of small vibrations about the given motion. The simplest
problem of this kind is that of the stability of a point of equilibrium,
which we shall discuss first. In this case, we can use the machinery of
tensor algebra developed in Chapter 10 to give an elegant method of solu-
tion for the small oscillations. A more general problem occurs when we
are given any particular solution to the equations of motion. We may
then ask whether that solution is stable, in the sense that every solution
which starts from initial conditions near enough to those of the given
solution will remain near that solution. This problem will be discussed in
Section 12-6. Methods of solution will be given for the special case of
steady motion.
12-1 Condition for stability near an equilibrium configuration. Let us
consider a mechanical system described by generalized coordinates
xi, . . . , Xf, and subject to forces derivable from a potential energy
V(xi, . . . , Xf) independent of time. If the system is subject to con-
straints, we will suppose the coordinates chosen such that xi, . . . , x/
are unconstrained. The coordinate system is to be fixed in time, therefore
the kinetic energy has the form
k.i=i *
Lagrange's equations then become
V-* d /-Mr ■ \ V" 1 dMlm * * j_ dV — n h — i f
1=1 l ' m=1 (12-2)
These equations have a solution corresponding to an equilibrium con-
figuration for which the coordinates all remain constant if they can be
solved when all velocity-dependent terms are set equal to zero. The
system can therefore be in equilibrium in any configuration for which
the generalized forces vanish:
|^=0, fc=l,...,/. (12-3)
dXk
473
474 THEORY OF SMALL VIBRATIONS [CHAP. 12
These / equations are to be solved for the equilibrium points, if any, of
the system.
The question of stability is easily answered in this case. If V(xi, . . . , x/)
is a minimum for an equilibrium configuration x\, . . . , Xy relative to all
nearby configurations x\ + Sxi, . . . , Xy + Sxy, then this is a stable
configuration. The total energy
E = T+V (12H1)
is constant. Let
E = V(x\, ...,xf) + iE (12-5)
be the energy corresponding to any initial conditions x\ + 5x\, . . . ,
Xf + 5x°; xi, . . . ,x/ near equilibrium. Then if 5x°, . . . , Sx 1 }; x°, . . . , x°
are small enough, we can make BE as small as we please. Since T is never
negative, the motion is restricted by Eq. (12^) to a region in the configura-
tion space for which
V(x u ...,*/)< V(xl ■ ■ ■ , xf) + SE. (12-6)
Since V is a minimum at (#?, . . . , x°), if 8E is sufficiently small, the
motion is restricted to a small region near x\, . . . , x/. Furthermore,
since
T < SE, (12-7)
the velocities x 1( . . . , xj are limited to small values. Therefore the equi-
librium is stable in the sense that motions at small velocities near the
equilibrium configuration remain near the equilibrium configuration.
Conversely, if V is not a minimum near xi, . . . , x°, then it is plausible
that the equilibrium is unstable, because in some direction away from
x° u . . . , x°, V will decrease. If we can choose the coordinates so that x\,
say, corresponds to that direction, and so that X\ is orthogonal to the
other coordinates, then Eq. (12-2) for Xi is
5 <*«*i> - JC 2 -axT XlXm = - SI • (12 " 8)
For small enough velocities such that quadratic terms in the velocities are
negligible, this becomes
M lA = -g- (12-9)
But as we move away from equilibrium in the x i -direction, dV/dxi be-
comes negative, and x x has a positive acceleration away from the equi-
librium point. In Section 12-3 we shall present a more rigorous proof
that the equilibrium is unstable if V is not a minimum there.
12-2]
LINEARIZED EQUATIONS OP MOTION
475
*?
Here the test for a minimum point should be recalled. If x\, .
is an equilibrium configuration for which Eq. (12-3) holds, then it is a
minimum of V(xi, . . . , x/) relative to nearby configurations, provided
that all the determinants in the following sequence are positive:
d V n
dxl
d 2 V
dxl
d 2 V
dxidXf
d 2 V d 2 V
dXzdXi
dx%
> 0,
where the derivatives are evaluated at xl,
d 2 V
dx\
d 2 V
dx^Xf
d 2 v
d 2 V
dx/dxi
dxj
.,*?.*
> o,
(12-10)
12-2 Linearized equations of motion near an equilibrium configuration.
We wish now to study the motion of a system in the neighborhood of an
equilibrium configuration. The coordinates will be chosen so that the
equilibrium configuration lies at the origin x x = • • • = x/ = 0. The
potential energy V is to be expanded in a Taylor series in x x , . . . , x/.
The constant term 7(0, . . . , 0) may be omitted as it does not enter
into the equations of motion. The linear terms are absent, in view of
Eqs. (12-3). If our study is restricted to small values of x u . . . , x/, we
may neglect cubic and higher-order terms in x u . . . , x { , so that
V = ^2 \ K 1d X ^h
k,l=l
where
dXkdx i/xi = . . . = xf = o
Since the coordinate system is stationary, the kinetic energy is
T = ]T) %M k ix k ±i.
k,l=l
(12-11)
(12-12)
(12-13)
In general, the coefficients M M may be functions of the coordinates,
but since the velocities are to be small, to second order in x u . . . , x f ;
x u . . . , if we may take Mm to be the values of the coefficients at
Xl = • • • = Xf = 0.
Equations (12-11) and (12-13) can be written in a suggestive way by
introducing in the /-dimensional configuration space a configuration
vector x with components x\, . . . , xf.
(*i,
,Xf).
(12-14)
1 W. F. Osgood, Advanced Calculus, New York: MacMillan, 1925, p. 179.
476 THEORY OF SMALL VIBRATIONS [CHAP. 12
The coefficients Km and Mm become the components of tensors
fK 11 ---K 1/ \
K =
K ff /
(12-15)
M lf \
M
These tensors are symmetric, or can be taken as such, since by Eq. (12-12)
Kki = K lk> (12-16)
and in the denning equation (12-13) only the sum i(Mki + Mi k ) is de-
fined as the coefficient of x k ±i = xi± k . Therefore, we may require that
M hl = M lk . (12-17)
The kinetic and potential energies may now be written as
T = £i • M • x, (12-18)
V = ix K x. (12-19)
The Lagrange equations (9-79) may be written as
M • x + K • x = 0. (12-20)
This equation bears a formal resemblance to Eq. (2-84) for the simple
harmonic oscillator. If we write Eq. (12-20) in terms of components,
we obtain a direct generalization of Eqs. (4-135) and (4-136) for two
coupled harmonic oscillators.
We may solve Eq. (12-20) by the same method used to solve Eqs.
(4-135) and (4-136). We try
x = Ce pt , (12-21)
where C = (C x , . . . , C/) is a constant vector whose components C\, . . . ,C/
may be complex. We substitute in Eq. (12-20) and divide by e pt :
p 2 M • C + K • C = 0. (12-22)
If we write this in terms of components, we obtain
/
2 iV 2 M kl + K kl )Ci = 0, * = 1, ...,/. (12-23)
12-3] NORMAL MODES OF VIBRATION 477
If Ci, . . . , Cf are not all zero, the determinant of the coefficients must
vanish:
p 2 M fl + X/i • • • v 2 M ff + K ff
= 0. (12-24)
This is an equation of order / in p 2 whose / solutions, p 2 = —u 2 , give
the / normal frequencies of oscillation. We may then substitute any pj
in Eqs. (12-23) and solve for the components Cjy of the vector Cy (except
for an arbitrary factor). The solution may then be obtained as a super-
position of normal vibrations, just as in Section 4-10 for two coupled
oscillators. In the next section we shall consider an alternative way,
utilizing the methods of tensor algebra developed in Chapter 10, of de-
termining the same solution.
12-3 Normal modes of vibration. If the coordinates x u . . . , x f are
orthogonal, the tensor M will be in diagonal form:
In = M k S kl . (12-25)
If the coordinates are not orthogonal, we can diagonalize M by the method
of Section 10-4, generalized to / dimensions. (We will use the same
method below to diagonalize the potential energy.) Let us suppose that
this has been done, and that the coordinates x u ...,x/ are the com-
ponents of x along the principal axes of M, so that Eq. (12-25) holds.
(If xi, . . . , x f are rectangular coordinates of a set of particles, M k is the
mass of the particle whose coordinate is x k .)
We now define a new vector y with coordinates yi, ■ ■ ■ ,y/ given by
y k =(M k x k yi 2 , k=l,...,f. (12-26)
Note that the configuration of the system is specified now by a vector y
in a new vector space related to the z-space by a stretch or compression
along each axis, as given by Eq. (12-26). The kinetic energy in terms
of y is
r=iy-y= £&2. (12-27)
Clearly, the expression for the kinetic energy does not change if we rotate
the ^/-coordinate system, which is our reason for introducing the vector y.
The potential energy is given by
v = y ■ W • y = X) Wkiym, (i 2 -28)
k,l=l
478
where
THEORY OF SMALL VIBRATIONS
W kl =
Ki
hi
Mi/zjlfj 1 / 2
The equations of motion are
y + Wy
[chap. 12
(12-29)
(12-30)
The tensor W is symmetric, and can therefore be diagonalized by the
method given in Section 10-4. Let ey be an eigenvector of W correspond-
ing to the eigenvalue W ' y.
W ey = Wfij. (12-31)
Let aij be the components of ey in the ^-coordinate system:
ey = (a lh . . . , a f j), j = 1, . . . ,/.
(12-32)
Then we may write Eq. (12-31) in terms of components in a form corre-
sponding to Eqs. (10-108) :
£ (W ki - Wj S k i)aij = 0, k = 1,
,/•
(12-33)
Again, the condition for a nonzero solution is
Wu - Wj W 12
W 21 W 22 - Wj ■■■
W n
W f2
W 2f
W ff - Wj
0. (12-34)
This is an algebraic equation of order / to be solved for the / roots Wj.
Note that it is the same as Eq. (12-24) if p 2 = —Wj and we divide the left
side of Eq. (12-24) by M x • M 2 ■ ■ • M t , remembering that M k i is now
given by Eq. (12-25). Each root Wj is to be substituted in Eq. (12-33),
which may then be solved for the ratios aiji a 2i : ■ • • : a/j. The aij can
then be determined so that ey is a unit vector:
i=i
(12-35)
The proofs given in Section 10-4 can be extended to spaces of any number
of dimensions, so we know that the roots Wj are real, and therefore the
12-3] NORMAL MODES OF VIBRATION 479
coefficients aij are also real. Moreover, the unit vectors ej, e* are orthog-
onal* for Wj 7* Wi. We have therefore
e y • e r = d ]r , (12-36)
or
/
X) ai i aiT = 8 *- (12-37)
In the case of degeneracy, when two or more roots Wj are equal, we
can still choose the aij so that the corresponding ej are orthogonal. The
situation is precisely analogous to that described in Section 10-4, except
that for / > 3 it cannot be visualized geometrically. The proof of
lemma (10-125) can be generalized to multiple degeneracies in spaces of
any number of dimensions.
Now let the components of the configuration vector y along e u . . . ,e/
beq u ..., qf.
y = £ «*;• < 12 - 38 )
In terms of components in the original y-coordinate system,
/
Vk = £ a kj qj. (12-39)
i=i
Conversely, by dotting e r into Eq. (12-38) and using Eqs. (12-32) and
(12-36), we obtain:
q r = X) a kr y k . (12-40)
it=i
These equations are analogous to Eqs. (10-67) and (10-69).
The potential energy in the coordinate system q\,...,qj, which
diagonalizes W, is
V=t, Will d 2 - 41 )
i=i
Since V is a minimum at the origin y = 0, the eigenvalues W u ...,W f
must all be positive; otherwise for some values of qi, . . . , q/, V would be
negative. If V were not a minimum, some of the eigenvalues Wj would
be negative. (The special case Wj = may or may not correspond to
* It is customary to use the term "orthogonal" rather than "perpendicular"
in abstract vector algebra when the vectors have only an algebraic, and not
necessarily a geometric, significance.
480 THEORY OF SMALL VIBRATIONS [CHAP. 12
a minimum, depending on higher-order terms which we have neglected.)
Let us set
Wj = a) . (12-42)
The kinetic energy (12-27) in this case is
T = £ Ml (12-43)
In view of Eqs. (12-41) and (12-43), the Lagrange equations separate
into equations for each coordinate qj\
Qi + <*hi=0, j=h...,f. (12-44)
The coordinates q, are called the normal coordinates. The solution is
q, — A, cos (a jt + Bj sin afi, j = 1, . . . , /, (12-45)
where Aj, Bj are arbitrary constants. We may write the solution in
terms of the original coordinates, using Eqs. (12-26) and (12-39) :
/
x h = M k 112 ^ a kj {Aj cos ufl + Bj sin ojjt). (12-46)
3=1
The coefficients are
Aj = qj (0) = X a k jMl l2 x k (0) (12-47)
and
/
Bj = «r X 4y(0) = 22 o>T 1 akiMl"± k (0). (12-48)
We therefore have the complete solution for small vibrations about a
point of stable equilibrium.
When the number of degrees of freedom is large, solving Eq. (12-34)
may be a formidable job which, in general, can be done only numerically
for numerical values of the coefficients. However, in some cases we may
know some of the roots beforehand (often we know that certain normal
frequencies are zero), or from symmetry considerations we may know
that certain roots are equal. Any such information helps in factoring
Eq. (12-34).
If V is not a minimum at x\ = • • • = x/ = 0, and some of the co-
efficients Wj are negative, then we obtain exponential-type solutions.
This proves that the motion is unstable in this case, since the solution
(except for very special initial conditions) will contain terms which in-
crease exponentially with time, at least until the linear approximation
12-4] FORCED VIBRATIONS 481
we have made in the equations of motion is no longer valid. The case
when some Wj is zero will not be discussed in detail here. In the linear
approximation we are making, the corresponding qj is constant in that
case, and this corresponds to what was called neutral equilibrium in
Chapter 2. The motion will proceed at constant fa until qj is large enough
so that nonlinear terms in qj must be considered.
It may be noted that in finding the normal coordinates we have found
a transformation from coordinates x\, . . . , x; to qi, . . . , <?/ which simul-
taneously diagonalizes two tensors M and K, or more correctly, which
simultaneously diagonalizes two quadratic forms, T and V. Unless two
tensors have the same principal axes, it is of course impossible simul-
taneously to diagonalize them by a rotation of the coordinate system.
However, if the coordinate system is allowed to stretch or compress along
chosen axes, as in the transformation (12-26), then we can bring two
quadratic expressions to diagonal form simultaneously (provided that at
least one is positive or negative definite). We first find the principal axes
of the first tensor. By stretching and compressing along the principal
axes, we can reduce this to a constant tensor (provided that the eigen-
values are all positive or all negative). In the case above, we reduced
M to 1 with the transformation (12-26). Since all axes are principal axes
for a constant tensor, the principal axes of the second tensor, as modified
by the stretching of coordinates, will reduce both tensors to diagonal form.
The reader will find it instructive to give a geometrical interpretation of
this procedure, in the case of tensors in two or three dimensions, by
representing each tensor by its associated quadric curve or surface, just
as the inertia tensor was represented in Section 10-5 by the inertia ellip-
soid. When we are dealing with vectors and tensors in physical space,
we ordinarily do not consider nonuniform stretching of axes because
this distorts the geometry of the space. When we deal with an abstract
vector space, we may consider any transformation which is convenient
for the algebraic purpose at hand.
»
12-4 Forced vibrations. We now wish to determine the motion of the
system considered in the preceding section when it is subject to prescribed
external forces F x (0, . . . , F/(t) acting on the coordinates x\, . . . , x f . We
will again restrict our consideration to motions which remain close enough
to the equilibrium configuration so that only linear terms in x\, . . . , x/
need to be included in the equations of motion. If we introduce the
vector
F(0 = (F u ..., F f ), (12-49)
we may write the equations of motion in the abbreviated form,
M-x-|-IC-x = F(fl. (12-50)
482 THEORY OF SMALL VIBRATIONS [CHAP. 12
where we have simply added the forces F(<) to Eq. (12-20). Note that
Eq. (12-50) may be obtained from the Lagrangian function
L = T - V - V, (12-51)
where T and V are given by Eqs. (12-11) and (12-13), and
/
V = - 22 ****(0- (12-52)
k=i
Again suppose that the coordinates xi, . . . , x/ are chosen to be orthog-
onal so that M is diagonal. If the coordinates x k are not initially orthog-
onal, and a rotation of the coordinate system is performed to principal
axes of M, then the components F k (t) must be subject to the same trans-
formation as the coordinates x k . Since we shall follow this process through
in the case where we diagonalize the tensor K, we shall not follow it through
in detail for M, but simply assume that, if necessary, it has been carried
out and that M is diagonal.
We transform now to the normal coordinates found in the preceding
section [Eqs. (12-26), (12-39), and (12-40)]:
x k - J2 Mk ll2 a kj qj, (12-53)
3-1
q 3 - = J2 Ml l2 a kj x k . (12-54)
k=l
The generalized forces Qj{t) associated with F k {t) are obtained by using
Eq. (9-30).
Qi{t) = X) Mjr lla a kJ F k ®. (12-55)
The inverse transformation j#
• /
F k (t) = J2 MV 2 a kj Qj{t), (12-56)
;=i
The reader may also verify Eqs. (12-55) by substituting Eqs. (12-53) in
Eq. (12-52) and calculating
dV
Qi=--^~- (12-57)
In normal coordinates,
V'=-J2 «i<?iW. (12-58)
3=1
12-4] FORCED VIBRATIONS 483
so that the equations of motion are
& + <*hi = Qi(0, 3 = 1, • • • . / • (12-59)
Each of these equations is identical in form with Eq. (2-86) for the un-
damped forced harmonic oscillator (b = 0). Therefore the normal modes
behave like independent forced oscillators, and the solution can be ob-
tained by the methods described in Chapter 2.
It is tempting to try to generalize our results to the case when linear
damping forces are also present. We can easily write down the appropriate
equations. In the general case when the coordinates are not orthogonal
and there is frictional coupling between coordinates, the equations of mo-
tion will be
X) (M k tfi + B ht ki + K klXl ) = 0, * = 1, ...,/, (12-60)
or, in vector form,
Mx+Bx+Kx=0. (12-61)
Unfortunately, as the reader may perhaps convince himself with some
experimentation, it is generally not possible simultaneously to diagonalize
three tensors M, B, K with any linear transformation of coordinates, even
if stretching is allowed, The method of the preceding section therefore
fails in this case, and there are no normal coordinates. The situation is
not improved by assuming that X\, . . . , x/ are orthogonal so that M is
diagonal, or even by assuming that there is no frictional coupling so that
B is diagonal. If we apply the transformations (12-53) and (12-54) which
diagonalize T and V, the coordinates q, are in general still coupled by
frictional forces:
9i + L Mr + o>fa = 0> C 12 - 62 )
r=l
where
b yr = £ M k ll2 Mr ll2 a kJ a lr B hl . (12-63)
k,l=l
Note that the matrix bj r is not diagonal even if B k i is. There is a special
case which sometimes occurs when the frictional forces are proportional
to the masses, so that B = 2TM. The method of Section 12-3 then works,
since in the y-coordinate system in which M — ► 1, we have B — > 271
and the normal coordinates q, along the principal axes of W satisfy the
separated equations:
qj + 27ft + o>hi = 0- C 12 " 64 )
484 THEORY OF SMALL VIBRATIONS [CHAP. 12
It may, of course, also happen that in the y-space in which M becomes 1
the transformed tensors B and K have the same principal axes, but this
would be an unlikely accident. When the damping forces are very small,
a perturbation method similar to that which will be developed in the
next section can be applied to find an approximate solution in terms of
damped normal modes.
Except in these special cases, the problem of damped vibrations can
be handled only by direct substitution of a trial solution like (12-21) in
the equations of motion (12-60). The secular equation analogous to
Eq. (12-24) is then of order 2/ in p. Each root allows a solution for the
vector C. If there are complex roots, they occur in conjugate pairs, p, p*,
with corresponding conjugate vectors C, C*. The two solutions (12-21)
can then be combined to yield a real solution which will be damped and
oscillatory, and can be called a normal mode. If all 2/ solutions are com-
bined with appropriate arbitrary constants, the general solution to Eqs.
(12-60) can be written. It is clear on physical grounds, since the frictional
forces reduce the energy of the system, that the real parts of all roots p
must be negative if F(x) has a minimum at x = 0; the mathematical
proof of this statement is a difficult exercise in algebra.
12-5 Perturbation theory. It may happen that the potential energy is
given by
V = V° + V, (12-65)
where V°(xi, . . . , x/) is a potential energy for which we can solve the
problem of small vibrations about a minimum point X\ = • • • = x/ = 0,
and where V'(xi, . . . , xj) is very small for small values of x\, . . . , x/.
We will call V° the unperturbed potential energy, and V the perturba-
tion. We expect that the solutions for the potential energy V will approxi-
mate those for the unperturbed problem. In this section we shall develop
an approximate method of solution based on this idea.
We shall assume that V is stationary at Xi = • • • = x; = 0, so that
f^) = 0. (12-66)
yXh / x l ^=,...=Xf=0
If this is not' the case, it is not difficult to find approximately the values
of x\, ... , Xf for which V is stationary. We leave this as an exercise.
The origin of coordinates should then be shifted slightly to x\, . . . , x°.
This will alter slightly the quadratic terms in V°, but these small changes
can be included in V. In any case, we will therefore have an expansion
12-5] PERTURBATION THEORY 485
of V around the equilibrium point of the form (12-65), with
V° = X) K*Wz, (12-67)
k,l
V = X) KiWi. (12-68)
k,l
where the coefficients K'u are small. The precise criteria that must be
satisfied in order that K' k i can be considered small will be developed as we
proceed.
We first transform to the normal coordinates g?, . . . , gy for the un-
perturbed problem. We then have
V° = £ i^feV, (12-69)
3=1
V = £ i^>2°9r°, (12-70)
T^> = X) M^MT^WmrUu (12-71)
ft, j
where TFy are the roots of the secular determinant (12-34) for the un-
perturbed problem, and where again we assume, for simplicity, that
Xi, . . . ,'Xf are orthogonal coordinates. The coefficients W'j r are to be
treated as small. The superscript "°" will remind us that the variables
q'j are normal coordinates for the unperturbed problem.
The equations of motion for q u . . . , q, are
$ + Whl + E W' jr q°r =0, j = 1, • • • , /• (12-72)
r=l
We see that the diagonal element of W adds to the coefficient of g°, while
the off -diagonal elements couple the unperturbed normal modes. We
expect that if W is small, there will be a normal mode of the perturbed
problem close to each normal mode of the unperturbed problem, that is,
a solution with frequency uij near o>° = (W?) 1/2 and for which q° is large
while the remaining q°, r ^ j, are small. However, if the tensor W° has
degenerate eigenvalues, so that two or more of the unperturbed fre-
quencies are equal (or perhaps nearly equal), then we expect that even a
small amount of coupling can radically change the motion, as in the case
of two coupled oscillators that we worked out in Chapter 4. This insight
will help in developing a perturbation method.
486 THEORY OF SMALL VIBRATIONS [CHAP. 12
If we try to find a normal mode of oscillation by substituting
<Z? = (V", j=l,...,f, (12-73)
in Eqs. (12-72), we obtain
(p 2 + WfiCj + X) W' jr Cr = 0. (12-74)
r=l
Now let us assume that the mode which we seek is close to some un-
perturbed mode, say j = 1. We then set
p 2 = -Wl - W'i,
Ci=l+ C' lt (12-75)
Cj = Cj, j = 2, . . . ,/,
where, if W[, C[, . . . , C} are zero, Eq. (12-73) represents a solution of the
unperturbed problem. Hence for the perturbed problem, we assume that
W' lt C[,...,C' f are small. We substitute Eqs. (12-75) in Eqs. (12-74),
and collect second-order terms on the right-hand side:
-W\ + WU = - X) w 'irC'r + WVJ'u (12-76)
r=l
(W> - Wlffl + W'n = - Y, W 'ir C 'r + W'lC'j, j = 2, . . . ,/.
r-1 (12-77)
When we neglect second-order terms, the first equation gives W[ :
Wi = W'n. (12-78)
Therefore
w f = -p 2 = W\ + W'n- (12-79)
Equations (12-77), if we neglect the right members, yield the coefficients
Cj:
C '^W^- i = 2 '■ <12 - 80)
The coefficient C[ is not determined; this corresponds to the fact that the
normal mode (12-73) may have an arbitrary amplitude (and phase),
although it must, of course, be near the amplitude (and phase) C? = 1
which was chosen in Eqs. (12-75) for the unperturbed solution. It will be
convenient to require that C\, . . . , C/ be the coefficients of a unit vector:
/
J2 C] = 1. (12-81)
3=1
12-5] PERTURBATION THEORY 487
We then obtain the following equation for C{ :
d = ~i £ (C'j) 2 . (12-82)
3=1
To first order in small quantities,
C\ = 0. (12-83)
By substituting in Eqs. (12-73), multiplying by an aribitrary constant
\Ae i *, and superposing the complex conjugate solution, we obtain the
first-order approximations to the perturbed normal mode:
ql = A cos (wi< + ff),
(12-84)
g° ~ t W ' n o cos ("i* + ®> j = 2,...,f,
where o>i is given by Eq. (12-79). We see that the first-order effect of the
perturbation is to shift u>\ by the diagonal perturbation coefficient W[\
and to excite the other unperturbed modes weakly with an amplitude that
is proportional to the perturbation coupling coefficients W'j\ and in-
versely proportional to the differences in unperturbed normal frequencies
(squared). This is a physically reasonable result.
We can now formulate more precisely the requirement that W be small.
In our derivation, we have assumed that
W{ « \W°i - TT?|, i = 2, ...,/, . (12-85)
C'j « 1. (12-86)
Equations (12-78) and (12-80) show that this is justified if
W' n « \W°j - W\\, j = 1, ■ • • , /• (12-87)
This is the condition for the validity of formulas (12-79) and (12-84).
First-order approximations to the remaining modes are obtained from
these formulas by interchanging the subscript «i» with any other.
The astute reader will note that the condition (12-87) in the diagonal co-
efficient W{\ is necessary only because we have neglected the last term in
Eqs. (12-77). Equations (12-77) are easily solved for C'j even if the last term
on the right is included. This allows us to remove the restriction on the size of
the diagonal coefficients if we wish. This is also obvious because we can always
include any diagonal coefficient in 7° [Eq. (12-69)]. The normal coordinates
488 THEORY OF SMALL VIBRATIONS [CHAP. 12
for the unperturbed problem are still the same; only the (squared) frequencies
are altered by adding additional diagonal terms. However, in the transformation
to normal coordinates, diagonal and off-diagonal terms become intermixed, so
that unless all terms in V (xi, . . . , x/) are small the off-diagonal terms of
V (<??> • • • > 9/) are unlikely all to be small.
Conditions (12-87) clearly cannot be satisfied if there is a degeneracy —
if, for example, W° = W 2 = W%. In that case, as mentioned earlier, we
expect that even with very small coupling of the unperturbed modes,
any perturbed mode with w 2 near W\ will show appreciable excitation of
all three unperturbed modes. We therefore set
p 2 = -w\ - W, (12-88)
and assume that only C 4 , . . . , C/ are small, while Ci, C 2 , C 3 may all be of
order 1. We substitute in Eqs. (12-74) and transpose second-order terms
to the right members:
/
(W' 11 -W')C 1 + W' 12 C 2 + W' 13 C 3 =-^W' lr Cr,
r=4
W' 21 Ci + (W 22 - W')C 2 + w 23 c 3 = - X) w 2r c r ,
r=4
W'atd + W' 32 C 2 + (W 33 - W')C 3 = - X) W'sfCr,
r= 4
(12-89)
and
(Wj - Wt)Cj + J2 W 'i£r = - X) W 'ir C * + W ' C i> 3 = 4, . . . , /.
(12-90)
If we neglect the right members, then Eqs. (12-89) become a standard
three-dimensional eigenvalue problem for the eigenvalue W and the
associated eigenvector (C 1; C 2) C 3 ). There will be three solutions corre-
sponding to three perturbed normal modes with frequencies w 2 =
W\ + W near the degenerate unperturbed frequency. In general, the
three roots W will be different, and so the perturbed modes will no longer
be degenerate. The remaining coefficients C4, . . . , C/ can be found to a
first-order approximation from Eqs. (12-90) by neglecting the right mem-
bers. We may also require that C be a unit vector [Eq. (12-81)]. In
analogy with Eq. (12-83), this will mean that to first order the three-
dimensional vector (Ci, C 2 , C 3 ) should be a unit vector. The case of double
12-5] PERTURBATION THEORY 489
or multiple degeneracy of any order must be treated in the same way.
Clearly, if the degeneracy is of high order, the first-order perturbation
equations [Eqs. (12-89) with right members zero] may be almost as
difficult to solve as the exact equations (12-74). When / > 4, we may
have more than one degenerate normal frequency; in that case, the above
method can be applied separately to each group of degenerate unper-
turbed modes to find the perturbed modes.
In cases of approximate degeneracy (W? = W 2 = W 3 ) when condi-
tions (12-87) fail for a group of neighboring unperturbed modes, the
method of the previous paragraph can also be applied. Equations (12-89)
are slightly modified by the addition of small terms, like W% — W°, in
the diagonal coefficients. The reader can readily formulate the procedure
for himself.
When the first-order approximate solution has been found, the ap-
proximate values of the coefficients W[, C'j may be substituted in the
right members of Eqs. (12-76), (12-77) [or Eqs. (12-89), (12-90)]. The
resulting equations are then solved to find a second-order approximation.
If, for example, we substitute Eqs. (12-80) in Eq. (12-76), we obtain the
second-order approximation to the frequency correction:
r=2 W 1 — W r
where we have used the fact that W is a symmetric tensor. We see that
the second-order frequency shift in mode 1 contains a contribution due to
coupling with each of the other modes. The modes tend to repel one
another in second order; that is, each higher frequency mode (W° T > W°)
reduces the frequency of mode 1, and each lower frequency mode increases
it. The same result was observed in the solution to the problem of two
coupled oscillators in Chapter 4. The procedure can be carried out in a
straightforward way to successively higher-order approximations, but the
labor involved rapidly increases.
To any order of approximation, we may introduce normal coordinates
qi, . . . , q/ for the perturbed problem by setting
gP = £ CjrQr, (12"92)
r=l
where Cj r , j = 1, ...,/, are the coefficients for the rth perturbed normal
mode found to any order of approximation by the perturbation theory.
The vectors C r — {C lr , . . . , C fr ) are, to the given order of approximation,
orthogonal unit vectors (or can be made so), as we shall see presently.
490 THEORY OF SMALL VIBRATIONS [CHAP. 12
We may therefore solve Eqs. (12-92) for
«r = E Crtl (12-93)
3=1
From Eqs. (12-92) and (12-73), if p 2 = -W° — W' r = — a? is the
approximate value for the frequency, then the approximate solution for
q r must be
q r = A r cos o) r t + B r sin u r t. (12-94)
Comparison with Eqs. (12-45) shows that the q r are (approximate) normal
coordinates. The Lagrangian, to the given order of approximation, must
therefore be
L = E (44? - Wa?), (12-95)
r=l
as may also be verified by straightforward substitution of Eqs. (12-92)
in Eqs. (12-69), (12-70), and (12-43), to any order of approximation
in Cj r . Alternatively, we may note that Eqs. (12-74) are just the equa-
tionsjs#e would obtain if we were to look for an eigenvector C of W =
W° -f- W corresponding to the eigenvalue — p 2 . Hence the approximate
solutions we have obtained for Eqs. (12-74) are also approximate solu-
tions to the problem of diagonalizing W. Equations (12-92) must there-
fore define approximate normal coordinates for the perturbed motion.
12-6 Small vibrations about steady motion. Let a mechanical system
be described by coordinates X\, . . . , Xf, and by a Lagrangian function
L(xi, . . . ,Xf) ±i, . . . , ±f) t). If a solution x\{t), . . . , x%t) is known,
we may look for solutions close to the known solution by defining new
coordinates y\, . . . ,yf\
x k = x° h (t) + Vk, h = 1, . . . ,/. (12-96)
We substitute in the Lagrangian L, and expand in powers of y u . . . y t ;
2/i, ... , yj. Since x\{t), . . . , x%t) satisfy the equations of motion, the
reader can readily show that no linear terms in y\, . . . , y/; y\, . . . , y/
occur in L. Terms in L independent of Vi, ■ . . ,y/',yi, . . • ,y/ do not
affect the equations of motion and may be ignored. If we assume that
Vu • • • tVf'iVi) • • ■ tVf are small, and that we may neglect cubic and
higher powers of small quantities, L becomes a quadratic function of the
new variables. The equations of motion will then be linear in y i} . . . , y/;
Vi, ■ ■ ■ , ilf', Vi, ■ ■ ■ , Vf- However, the coefficients in the equations will,
in general, be functions of the time t, and the methods we have so far de-
veloped will not suffice to solve them. To develop methods for solving
equations with time-varying coefficients is beyond the scope of this book.
12-6] SMALL VIBRATIONS ABOUT STEADY MOTION 491
We will therefore consider only cases when the coefficients in the linearized
equations turn out to be constant.
We can guarantee that the coefficients will be constant by restricting
ourselves to steady motions. Suppose that some of the coordinates
Xi, . . . , Xf are ignorable, i.e., do not appear in the Lagrangian function.
We will also assume that L does not depend explicitly on t. We define a
steady motion as one in which all of the nonignorable coordinates are
constant. This definition evidently depends upon the system of co-
ordinates chosen. We should perhaps define steady motion as motion for
which, in some coordinate system, the nonignorable coordinates are all
constant.
We have seen in Section 9-10 that ignorable coordinates are particularly
easy to handle in terms of the Hamiltonian equations of motion. Let us
therefore introduce coordinates x lt . . . , x f , and corresponding momenta
Pi, ■ • • ,Ps, of which xn+i, . . . , x/ are ignorable. The Hamiltonian
function is
H = H(x u . . . ,x N ;vu ■ ■ ■ , Pn, Pn+i, ■ ■ ■ , Pf). (12-97)
In view of Eqs. (9-198), the momenta Pn+u ■ ■ ■ ,Pf are all constant.
We have therefore to deal only with 2N equations (9-198), which for
steady motion reduce to
For given values Pn+i, . . . , p°, we are to find the solutions, if any, of
these equations for Xi, . . . , Xn',Pi, • • ■ , Pn- Any such solution a;?, . . . , x%;
Pi, • • - 1 Pn defines a steady motion. The ignorable coordinates will all
have constant velocities given by
*' ~ Uo'
N+l,...,f, (12-99)
where the subscript " " implies that the derivative is to be evaluated at
• £, i> • • • j ■*-#> Pi> • • • , Pf
Given a steady motion, let us choose the origin of the coordinate system
so that x\ = • • • — x% = pi = • ■ • '= p% = 0. In order to look for
motions near this steady motion we hold Pn+i, ■ • ■ ,Pf fixed and expand
H in powers of x\, . . . , xn, Pi, ... , Pn, which we regard as small. We
may omit any terms which do not depend on x lt . . . ,pn- Linear terms
are absent because of Eqs. (12-98). If we neglect cubic terms in small
quantities, H becomes a quadratic function of x i, ... , xn, Pi, ■ • • , Pn,
with constant coefficients. It may be that H separates into a positive
492 THEORY OF SMALL VIBEATIONS [CHAP. 12
definite "kinetic energy," ' T' {p lt ..., p N ), and a "potential energy,"
'V'(xi, . . . , x N ). In that case, the methods of the preceding sections are
applicable. In order to apply these methods, we must express the "kinetic
energy," 'T\ in terms of x u . . . , x N , which may be done by solving for
Pi, • ■ ■ i Vn the linear equations
x k = ^ , k=l,...,N. (12-100)
apk
The problem is thus reconverted to Lagrangian form, with 'L' {x u . . . , x N ;
±i, . . . , An) = 'T' — 'V. Note, however, that we cannot obtain the
correct 'L' simply by substituting x° from Eq. (12-99) in the original L.
The transition to Hamiltonian form is necessary in order to be able to
eliminate the ignorable coordinates from the problem by regarding
Pn+i, ■ ■ ■ , Pf as given constants. The "potential energy" 'V will con-
tain terms involving p N +i, ■■ ■ ,Pf frcm the original kinetic energy,
and these will appear with opposite sign in 'L'.
If l V'(xi, . . . , xn) has a minimum at x\ = • • • = x^ = 0, then
xi, . . . , xn, if they are small enough, undergo stable oscillations. We
may then say that the given steady motion is stable in the sense that for
nearby motions (with the same Pn+i, ■ ■ ■ , Pf), the coordinates oscillate
about their steady values. These oscillations can be described by normal
coordinates, which may be found by the method of Section 12-3.
If the coordinate system we use is a moving one, or if magnetic forces
are present and must be described by a velocity-dependent potential
(9-166), or if, as often happens, the ignorable coordinates are not orthog-
onal to the nonignorable coordinates, then cross product terms XkPi will
appear in H. Thus, in general, the quadratic terms in H have the form
JV
H = ^2 (\a k ipkpi + bkixkpi + ickiXkXi), (12-101)
k,l=l
where we may as well assume that
aik = a k i, cik = c k i. (12-102)
The coefficients a«, bu, Cki are functions of the constants Pn+i, ■ ■ ■ ,Pf
and of the particular steady motion whose stability is in question. We
no longer have a separation into kinetic and potential energies, and the
methods of the preceding sections can no longer be applied. It may be
that H as given by Eq. (12-101) is positive (or negative) definite in
x\, . . . ,xn, Pi, • ■ ■ ,Pn, in which case we may be sure that the steady
motion is stable. For x lt . . . , X&, p u . . . , Pn must remain on a surface
of constant H, and if H is positive definite, this surface will be an "ellip-
soid" in the 2iV-dimensional phase space.
12-6]
SMALL VIBRATIONS ABOUT STEADY MOTION
493
In any case, we may study the small vibrations about steady motion
by solving the linearized equations given by the Hamiltonian function
(12-101) :
Xk
N
i=i
*k ~ £ ( a kiPi + bi k x{),
N
Pk = — ^2 (bkiPi + Ckixi), k = 1, . . . , N.
(12-103)
We could return to a Lagrangian formulation involving N second-order
equations in x 1} . . . , xn, but it is just as easy to deal directly with
Eqs. (12-103). Let us look for a normal mode in which all quantities have
the same time dependence:
x k = X k e pt , p k = P k e pt . (12-104)
We substitute in Eqs. (12-103) and obtain the 2N linear equations
N
X) K&» - V hi)Xi + a k iPi] = 0,
2=1
X) [^iXi +(hi + p S k i)Pi] = 0, k = 1, . . . , N. (12-105)
i=i
The determinant of the coefficients must vanish:
0-2N
blN
b 2 N
bNN + P
11 -
P
b 2 i
■ «ii
612
622 —
p. .
• <*21
en
c 12
•&11 + P
C 2 1
C22
■
2>21
CiVl
CJV2
bNi
0. (12-106)
We now note that if p is any root of this equation, so is —p. First, let us
set p = —p' in the determinant. Now, interchange the upper N rows
and the lower N rows. Next, interchange the left N columns with the
right N columns. Finally, interchange rows and columns, i.e., rotate
about the main diagonal. None of these operations changes the value
of the determinant (except possibly its sign, which does not matter).
We now have the same equation for p' that we originally had for p, so
that if p is a root, so is p'. We see therefore that when we expand the
determinant (12-106), only even powers of p appear, and we have an
algebraic equation of degree N in p 2 . If the roots are all negative, as they
494 THEORY OF SMALL VIBRATIONS [CHAP. 12
will be if H in Eq. (12-101) is positive definite, then the normal modes
are all stable. Each root p 2 = — u 2 gives two values p = ±iwy. We
substitute p = iwj in Eqs. (12-105) and solve for Xij, Pij, which will,
in general, be complex. There is, of course, an arbitrary constant which
may be chosen in any convenient way. The solutions for p — —iuj
will be X*j, P*j. We substitute in Eqs. (12-104), multiply by an arbitrary
constant Aje i>j and superpose the two complex conjugate solutions to
obtain the real solution for the normal mode j:
where
x k = AjCkj cos (wjt + hi + Oj),
p k = AjD kj cos (ujt + 7k, + Oj),
X kj = ^C kj e ih ',
Pkj = %D kj e iyki ,
(12-107)
(12-108)
and Aj and 6j are an arbitrary amplitude and phase. The general solution
is now a superposition of normal modes :
N
Xk
= J3 A i Ck i cos ( w ^ + Pv + e J')»
J'=l
iV
(12-109)
Pk = ^2 AjD ki cos {ujt + y k j + dj).
3=1
Note that we cannot represent the above result in terms of normal
coordinates q\, ■ ■ ■ ,qN linearly related to x-y, . . . , xn on account of the
phase differences f$ k j, 7kj which arise because of the cross terms in co-
ordinates and momenta. It is possible to find a linear transformation
connecting the 2N variables x 1; . . . , x^; pi, . . . , p^ with a set of normal
coordinates and momenta </i, . . . , <?#; pi, . . ■ , Pn, each of which oscil-
lates at the corresponding normal frequency. Such transformations be-
long to the theory of canonical transformations of Hamiltonian dynamics,
and are beyond the scope of this book.
If any-root p 2 of Eq. (12-106) is positive or complex, the corresponding
normal mode is unstable, and the coordinates and momenta move ex-
ponentially away from their steady values. A root p 2 = would corre-
spond to neutral stability. One common case in which roots p 2 =
arise occurs when an ignorable coordinate has been included among
the x i, . . . , xn. If Xj is ignorable, and is included in X\, . . . , x^, then
Pi is constant and may take any value. If we take pj slightly different
from p° for the initially given steady motion, there is a new steady
12-6] SMALL VIBRATIONS ABOUT STEADY MOTION 495
motion with x j constant and slightly different from xj. This new motion
is given by
Xj = Aj + BjXjt, all other x k = Ak, (12-110)
where Ak may be slightly different from x° k . This motion corresponds to
a normal mode with p 2 = 0. In some cases, the algebra required to ignore
explicitly a coordinate Xj is too formidable, and we may prefer to include xj
among the nonignorable coordinates. This increases the degree of the
secular equation (12-106) by one, but since the extra root is p 2 = 0, we
know that p 2 will factor out and the remaining equation will have the
same degree as if we had ignored Xj. It may also be that we .have chosen
a coordinate system in which some ignorable coordinate Xj does not ap-
pear. A root p 2 = of Eq. (12-106) will still occur. Since the coordinates
actually used will be functions of the ignorable one (among others), in
the corresponding normal mode, several or all of the coordinates may
exhibit constant velocities. These may be found by substituting in the
equations of motion (12-103).
The case of degeneracy, when a multiple root for p 2 occurs, is more
complicated for Eqs. (12-103) than for Eqs. (12-20), where the force
is derivable from a potential energy depending only on x\, . . . , x N . We
can no longer make use of the diagonalization theory for a symmetric
tensor to show that for a multiple root p 2 , Eqs. (12-105) have a corre-
sponding multiplicity of independent solutions, as we did for the analogous
equations (12-23) or (12-33). Sometimes Eqs. (12-105) may have only
one independent solution, even when p 2 is a multiple root of Eq. (12-106),
and we must look for other forms of solution than (12-104). We will not
carry out the algebraic details here,* but the result is that when
Eqs. (12-105) do not yield enough independent solutions for X k , Pk for
a multiple root p 2 , X k , Pk should be replaced in Eqs. (12-104) by poly-
nomials in t of degree (n — 1), where n is the multiplicity of the root p 2 .
The resulting expressions must be substituted in Eqs. (12-103), which
then give 2Nn relations between the 2Nn coefficients in the 2N poly-
nomials. These relations can be shown to leave just n arbitrary coefficients,
so that the correct number of arbitrary constants are available. Alter-
natively, we can slightly alter the coefficients in the Hamiltonian (12-101)
so that the degeneracy in p 2 is removed, find the solution, and then find
its limiting form as the coefficients approach their original values. (See
* For a further discussion of problems of small vibrations, see E. J. Routh,
Dynamics of a System of Rigid Bodies, Advanced Part. New York: Dover Pub-
lications, 1955. Chapter 6. A less complete but more elegant treatment using
matrix methods is given in R. Bellman, Stability Theory of Differential Equations,
New York: McGraw-Hill, 1953.
496 THEORY OF SMALL VIBRATIONS [CHAP. 12
Problem 24, Chapter 2.) When powers of t appear in the solution, it is
clear that the solution is not stable even when p 2 is real and negative, but
represents an oscillation whose amplitude after a long time will increase
as some power of t. Hence degeneracy generally implies instability in the
case of Eqs. (12-103). It will be found that multiple roots of Eq. (12-106)
ordinarily mark the boundary between real and complex solutions for
p 2 in the sense that a small change in some coefficient au, bu, or c k i
will split the degeneracy and lead on the one hand to two real, or on the
other hand to two complex, roots for p 2 , depending on the sense of the
change. The situation is closely analogous mathematically to the problem
of the damped harmonic oscillator, where a double root for p in Eq. (2-125)
marks the dividing line between the overdamped and underdamped cases
and leads to solutions linear in t. In the present case, there is no damp-
ing; Eq. (12-106) contains only even powers of p, and if complex con-
jugate roots for p 2 occur, then the corresponding four roots p have the
form ±7 ± io), and some of the solutions grow exponentially. Thus
multiple roots can mark the boundary between stable and unstable cases.
In the case of Eqs. (12-20), where the forces are not velocity-dependent,
the boundary between stability and instability occurs only when some
root for p 2 is zero; degenerate negative roots for p 2 always correspond to
stable solutions.
We should further remark that even when the roots p 2 are all negative
and distinct, so that the solutions of Eqs. (12-103) are all stable, we
cannot guarantee that the exact solutions of the nonlinear equations aris-
ing from the complete Hamiltonian (12-97) are stable. For vibrations
about an equilibrium point when the forces are derivable from a potential
energy, we were able to prove that stable solutions of the linearized equa-
tions are obtained only around a potential minimum, and that in that
case, the solutions are absolutely stable if the amplitude is small enough.
We saw above that if the Hamiltonian H is positive (or negative) definite
near the steady motion, then we can also show that the solutions are
stable. But the solutions of Eqs. (12-103) may all be stable even when
H is not positive or negative definite. In that case, all we can say is
that, for as long a time as we please, if we start with sufficiently small
amplitudes, the solutions of the exact problem given by the Hamiltonian
(12-97) will "approximate those of the linearized problem given by the
Hamiltonian (12-101). This is true because the nonlinear terms can be
made as small as we like by making the amplitude sufficiently small,
and then their effect on the motion can be appreciable, if at all, only if
they are integrated over a very long time. Nevertheless, if we neglect
the nonlinear terms, we are prevented from asserting complete stability
for all time. Cases are indeed known where the linearized solutions are
stable, and yet no matter how near the steady-state motion we begin,
12-7] BETATRON OSCILLATIONS IN AN ACCELERATOR 497
the exact solution eventually deviates from the steady-state motion by
a large amount. To find the criteria that determine ultimate stability
in the general case is perhaps the outstanding unsolved problem of
classical mechanics.
12-7 Betatron oscillations in an accelerator. In a circular particle ac-
celerator, for example a cyclotron, betatron, or synchrotron, charged
particles revolve in a magnetic guide field which holds them within a
circular vacuum chamber as they are accelerated. Since the particles
revolve many times while they are being accelerated, it is essential that
the orbits be stable. Since the particle gains only a small energy incre-
ment at each revolution, it is permissible to study first the stability of
the orbits at constant energy, and then to consider separately the ac-
celeration process itself. We will be concerned here only with the
stability problem at constant energy E. Let us assume that the mag-
netic field is symmetrical about a vertical axis, so that we may write,
using cylindrical polar coordinates (Fig. 3-22),
B(p, <p, z) = B z (p, z)k + B r {p, z)h. (12-111)
The magnetic field in a synchrotron or betatron is also a function of time,
increasing as the energy increases, but since we are treating E as constant,
we also take B as constant. We will suppose that in the median plane
z = 0, the field is entirely vertical :
B(/o, <p, 0) = B l0 (p)k. (12-112)
A particle of appropriate energy E may travel in a circle of constant radius
p = a(E). We call this orbit the equilibrium orbit. We are interested in
the stability of this orbit; that is, we want to know whether particles
near this orbit execute small vibrations about it. Such vibrations are
called betatron oscillations because the theory was first worked out for the
betatron.
The vector potential (Section 9-8) for a magnetic field with symmetry
about the z-axis can be taken to be entirely in the ^-direction:
A = A v {p, z)m, (12-113)
so that
B = V X A
= ^7<"^- h ^- (12 - 114)
498 THEORY OF SMALL VIBRATIONS [CHAP. 12
We see from Eq. (12-114) that A is given by
A,(p, z) = p- 1 fpB z (p, z) dp, (12-115)
Jo
since A v must vanish at p = because of the ambiguity in the direction
of m. Note that 2irpA v is the magnetic flux through a circle of radius p.
The Lagrangian function is given by Eqs. (9-154) and (9-166) :
L = im(p 2 + pV + z 2 ) + e - p?A v (p, z), (12-116)
where e, m are the charge and mass of the particle to be accelerated. If
the velocity of the particle is comparable with the speed of light, that is,
if the kinetic energy is comparable with or greater than mc 2 , the relativistic
form for the Lagrangian should be used (Problem 23, Chapter 9). The
momenta are
dL
Vp = _ = m p,
p„ = mp 2 <p + - pA v ,
(12-117)
p„ = mz.
The Hamiltonian function is given by Eq. (9-196) or (9-200) :
rr _ Pp 2 , P* , [P*> — (e/c)pA v ] 2
2m T 2m T 2mp2
(12-118)
We see that <p is ignorable and p 9 may be taken to be a given constant.
The Hamiltonian function then has the form
with
h = <r + 'v,
(12-119)
trp}
= ^ = 5^ + ^
(12-120)
'V
_ [Vv — (e/c)pA v ] 2
9mn2
(12-121)
The problem reduces to an equivalent problem of static equilibrium.
The steady motions are given by the solutions for p, z of the equations
d'V e
— = 1 pvB r = 0, (12-122)
d'V ( e \
— = -pip \m<p + - B A = 0, (12-123)
12-7] BETATRON OSCILLATIONS IN AN ACCELERATOR 499
where we have used Eqs. (12-117) and (12-114). The first equation
above is satisfied in the median plane z = 0, and usually nowhere else
(p<p 9* 0). The second equation gives
*=--L B °°( p) ' (12_124)
7/tC
which is equivalent to Eq. (3-299). We may solve Eq. (12-124) for p,
given <p, or alternatively, for </> with p = a, the radius of the equilibrium
orbit. Note that the energy %ma 2 <p 2 of a particle executing the steady
motion is less than the total energy H of the particle whose motion we are
studying, but the difference is of second order in small quantities for
vibrations near the steady motion. According to the development in the
preceding section, the two particles should be chosen to have the same p v .
We now set
p = a + x, (12-125)
where a is the radius of the orbit for the steady motion. Next, by expand-
ing in powers of x, z, x, z, we obtain
T = \m{x 2 + z 2 ), (12-126)
'V = 4m« 2 (l - n)x 2 + %ma 2 nz 2 , (12-127)
where we have set
co = ID = —
mc
ei^a) t (12-128)
mc
*=-(irir) ' ( 12 " 129)
and have used Eqs. (12-114), (12-117), (12-124), and
dB z
dp
dB„
dz
which follows from the fact that
V X B
=
(12-130)
(12-131)
as we can see from Eq. (9-162). The quantity n is called the field index.
We see immediately that the motion is stable only if
< n < 1. (12-132)
In a cyclotron, the field is nearly constant at the center, so that n « 1,
and then falls rapidly near the outside edge of the magnet. In a betatron
or synchrotron, the magnetic field has a constant value of n and increases
500 THEORY OF SMALL VIBRATIONS [CHAP. 12
in magnitude as the particles are accelerated so as to keep a constant. We
see from Eq. (12-129) that the value of n does not change as B z is increased
provided the shape of the magnetic field as a function of radius does
not change, that is, provided dBJdz increases in proportion to B z .
Since the variables x, z are separated in 'T' and 'V, we can immedi-
ately write down the betatron oscillation frequencies. It is convenient
to express them in terms of the numbers of betatron oscillations per
revolution, v x and v z :
Wx /-, \l/2
v x = — = (1 — n) ,
w (12-133)
w z 1/2
If there are imperfections in the accelerator, so that B z is not inde-
pendent of <p, the difference between B z and its average value gives rise
to a periodic force acting on the coordinate x. The resulting perturbation
of the orbit can be treated by solving the corresponding forced harmonic
oscillator equations. If B p is not zero everywhere in the median plane
(2 = 0), vertical forces act which drive the vertical betatron oscillations.
In general, such imperfections also lead to variations in the field index n,
so that n = n(<p), and n for the steady motion becomes a periodic func-
tion of time. In alternating gradient accelerators, the field index n is
deliberately made to vary periodically in azimuth <p. The solution of
this problem is too complex for inclusion here.
12-8 Stability of Lagrange's three bodies. A particular solution of
the problem of three bodies moving under their mutual gravitational
attractions was discovered by Lagrange. This solution is a steady motion
in which the three masses remain at the corners of an equilateral triangle
as they revolve around their common center of mass. We wish to in-
vestigate the stability of this steady motion. This problem is an example
of a rather general class of problems in celestial mechanics concerned
with the stability of particular solutions of the equations of motion.
When the particular solution is a steady motion, the problem can be
treated by the method of Section 12-6.
We will simplify the problem by considering only motions confined to
a single plane. There are then six coordinates, two for each particle. A
little study shows that there are three ignorable coordinates. Two of them
represent rigid translations of the three particles, and may be taken as
the cartesian coordinates of the center of mass. The corresponding con-
stant momenta are the components of the total linear momentum. The
third ignorable coordinate will represent a rigid rotation of the three
particles about the center of mass. The corresponding constant mo-
12-8]
STABILITY OF LAGRANGE'S THREE BODIES
501
mentum is the total angular momentum. The remaining three non-
ignorable coordinates will specify the relative positions of the three parti-
cles with respect to each other. These must be constant in a steady-
motion; therefore any steady motion must be a rigid translation and
rotation of the system of three bodies. We may, for example, choose
the coordinates as in Fig. 12-1. Here, X and Y are coordinates of the
center of mass, variation of a with the remaining coordinates held fixed
represents a rotation of the entire system about the center of mass, and
ri, r 2 , determine the shape and size of the triangle formed by the masses
mi, m 2 , m 3 . We expect to find three normal modes of vibration of r u r 2 , 6
about their steady values. If we should happen to overlook any ignorable
coordinate, we will find only those steady motions in which that co-
ordinate is constant. The ignorable coordinate will then reveal itself as
a zero root (p 2 = 0) of the secular equation (12-106).
According to Eq. (4-127), the kinetic energy will separate into a part
depending on X and Y, and a part depending on r\, r 2 , 6, and a. The
potential energy depends only on r u r 2 , and 0. The coordinates X, Y are
therefore orthogonal to r u r 2 , 6, and a, and the center-of-mass motion
separates out of the problem. The center-of-mass energy
T B
pf (x j + n
2M
(Px + Vy), , M = mi + m 2 + m 3 ,
(12-134)
is constant, and may be omitted from the Hamiltonian. The center of
mass moves with constant velocity, and we may study separately the
motion relative to the center of mass. The ignorable coordinate a is
evidently not orthogonal to 6, since the angular velocity of m x involves
Fig. 12-1. Coordinates for the three-body problem.
502 THEORY OF SMALL VIBRATIONS [CHAP. 12
(6 -\- a) and this appears squared in the kinetic energy. This is not an
accidental result of our choice of coordinates, but an inherent consequence
of the fact that rotation of the system as a whole influences the "internal"
motion described by r\, r 2 , 6.
It would be a straightforward algebraic exercise to set up the kinetic
energy in terms of ri, r 2 , 9, a, find the momenta p r iPr2, Pe, Pa, set up the
Hamiltonian, find the steady motions, and carry through the procedure
of Section 12-6 to find the secular equation (12-106), which would be a
third-order equation in p 2 whose roots determine the character of small
deviations from steady motion. This procedure, however, is extremely
tedious, as the reader may verify. It turns out, interestingly enough,
that a less laborious way of finding the actual solution is to abandon the
Hamilton-Lagrange formalism, and to set up the equations of motion
from first principles. We will still need the results of the above general
considerations as a guide to the solution. The algebra is still sufficiently
involved so that there is a rather high probability of algebraic mistakes.
It is therefore desirable to replace 0, a by the coordinates «i, a 2 shown in
Fig. 12-2, so as to introduce an algebraic symmetry between particles
mi and m 2 . This reduces the amount of algebra needed and provides a
check on the results, in that our formulas must exhibit the proper sym-
metry between subscripts "l" and "2". Neither a\ nor a 2 is ignorable now,
and we see that the ignorable coordinate no longer appears explicitly.
We could not make use of the ignorable property anyway, since we are
not going to write the equations in Hamiltonian form. Our secular equa-
tion will turn out to be of fourth order in p 2 , but we know that one root
will be p 2 — 0, and can be factored out. We show also in Fig. 12-2
several auxiliary variables r 3 , 6i, 6 2 , 03 which will be needed.
We will write the equations of motion of mi in terms of components
directed radially away from m% and perpendicular to the radius r\. In
applying Newton's laws of motion directly, we must refer all accelera-
tions to a coordinate system at rest. The acceleration of Wi is its acceler-
ation relative to wi 3 plus the acceleration of m 3 . The latter acceleration
can be found by applying Newton's law of motion to WI3, which is at-
m 3
Fig. 12-2. Alternative coordinates for the three-body problem.
12-8] STABILITY OF LAGRANGE'S THREE BODIES 503
tracted by m x and m 2 . The force on mi is the gravitational attraction of
m 2 and m 3 . We have therefore, in the radial direction,
/.. . 2 , miG . m 2 G . \ m x m 3 G m x m 2 G
mi I f i — rjai -\ l — + -~ cos 3 J = — cos X .
(12-135)
The corresponding equation for the motion of m lt perpendicular to r lt is
mi La, + 2nd! - ^ sin a ) = - M^ sin X . (12-136)
Two similar equations can be written for the motion of m 2 . The four
equations can be conveniently rewritten in the form:
.2 , (wii + m 3 )(? , m 2 G , m 2 G n
fi — rial + v ' 1 t- cos 3 H — cos 6i = 0,
r? r\ r\
(12-137)
.2 , (m 2 + m 3 )G , m-iG „ . mig „
r 2 — r 2 a 2 + y ' H 7- cos 3 H — cos 2 = 0,
r% r\ r%
, «. . m 2 G . „ . m 2 G . „ n
nan + 2r i a l t- sin 3 -\ — sin X = 0,
r 2 r 3
r 2 a 2 + 2r 2 a 2 H r- sin 3 — sin 2 = 0.
r\ r 3
The algebraic symmetry between subscripts "l" and "2" is exhibited in
the above equations. (Note that 3 = «i — «2 and changes sign if we
interchange the two particles.) The auxiliary variables X , 2 , 3 , r 3 may
be expressed in terms of r u r 2 , a u a 2 by using the sine and cosine laws
for the triangle.
We now see why it is easier to use Newton's laws directly here. We
can express the acceleration of m 3 very simply in terms of the gravitational
forces on m 3 . In the Lagrangian formulation of the corresponding equa-
tions for r lt r 2 , a lt a 2 (or a, 0), the terms which represent the acceleration
of m 3 have to be expressed kinematically, i.e., in terms of the coordinates,
velocities, and accelerations of m x and m 2 , because they are derived by
differentiation of the kinetic energy T in generalized coordinates. This is
very complicated, and involves explicitly the position of the center of
mass relative to m lt m 2 , m 3 , which we do not need in the present formula-
tion. The resulting equations are equivalent to Eqs. (12-137), but con-
siderably more complicated in form. One reason for the simplicity of
Eqs. (12-137) is, of course, the use of the auxiliary variables 6 U 2 , 3 , r 3 ;
504 THEORY OF SMALL VIBRATIONS [CHAP. 12
in the Lagrangian formulation, any such auxiliary variables would have
to be differentiated with respect to r lt r 2 , a lt a 2 in order to write down
the equations of motion.
We first look for steady motions. We know from our preliminary dis-
cussion that a steady motion can only be a rigid rotation about the center
of mass (plus a uniform translation). We therefore take r\, r 2 , r 3 , 0i, 2 , 03
to be constant, and set
a 2 = at, «i = ot + 63- (12-138)
If we substitute into the last of Eqs. (12-137), we obtain
\ sin 3 = A sin 02- (12-139)
From the law of sines,
rx r 3
(12-140)
sin 2 sin 3
we then have, unless sin 03 = sin 2 = 0,
r\ = rl. (12-141)
In the same way, from the third of Eqs. (12-137), we find r 2 = r 3 , unless
sin 3 = sin 0i = 0. We may therefore set
r\ = r 2 = r 3 = a,
- =, - * (12 " 142)
The only possible steady motion, unless the masses lie in a straight line,
is one in which the three masses lie at the corners of an equilateral triangle.
We must still verify that the first two of Eqs. (12-137) are satisfied.
This is the case if
a, 2 = ^> (12-143)
a 3
where M is the total mass. This is the particular solution of the three-
body problem found by Lagrange. The case when the three masses lie
in a straight line is left as an exercise.
We now seek solutions for motions near the steady motion. Let us set
r x = a + xi, r 2 = a + x 2 , (12-144)
«1 = Ut + ^7T + 6l, ol 2 = ut + e 2 ,
where xi, x 2 , e x , e 2 are four new independent variables which we will
12-8] STABILITY OF LAGRANGE'S THREE BODIES 505
regard as small. We substitute in Eqs. (12-137), retaining only linear
terms. We first calculate
03 = &r + «i - «2, (12-145)
and, to first order, from the law of cosines,
r % = a 2 [l + ^^1 + V3 (ei - «„)] • (12-146)
Now, from the law of sines,
sin 0i = ^sin0 3 ; (12-147)
f3
* = I - \ (€l - €2) - l^ 3 ^ 1 ' ( 12 " M8 >
Xi
—
X2
a
X!
—
X 2
hence, to first order,
and, similarly,
e 2 = § - \ (6 X - e 2 ) +\ V3^^- (12-149)
We note as a check that X + 2 + 03 = ir. We are ready to substitute
in Eqs. (12-137), which, to first order, become
„ . ( 2 , 2mi + 2m 3 — jm 2 r \
xi — 2owei — I w H ^3 O- J xi
9m 2 G 3\/3 m 2 G
4a3> ** 4a 2
(«i - « a ) = 0,
o • ^ 2 ■ 2m 2 + 2rre 3 — jmi „\
x 2 — 2ao>e 2 — I a> H ^ ^ I z 2
9miC? 3\/3 m x (? , . „
- -4^~ Xl 4^~ ^ - e »> = °'
.. , „ . 3\/3 m 2 G , s 9m 2 G , , _ n
atx + 2wi;i 4q3 (zi - s 2 ) ^j" («i — «2) = 0,
oe 2 + 2wi: 2 4q3 (xi - ^2) + -^p - (d — «2) - 0.
(12-150)
Note that the second and fourth equations may be obtained from the first
and third by interchanging the subscripts "l" and "2" and reversing the
signs of «!, a 2 , and o>; this symmetry follows from the choice of coordi-
nates in Fig. 12-2.
A normal mode is to found by setting
x x = X x e pt , x 2 = X 2 e pt ,
ci = Eie pt , e 2 = E 2 e
,* . - *.** ( 12 " 151 >
506 THEORY OF SMALL VIBRATIONS [CHAP. 12
For convenience we set
p = (G/a d ) l "P. (12-152)
Substituting Eqs. (12-151) in Eqs. (12-150) and using Eq. (12-143),
we obtain
(t>2 oar , 9 \Xi 9 ^2
- (^3. m2 + 2M 1/2 p) E,+^ m 2 E 2 = 0,
9 Xi , ( d2 ,,t , 9 \ X 2 3\/3 _
-4 mi ^ + ^ P -3M + i m 1 ) — -- r m 1 E 1
+ (^m 1 -2M 1 ' 2 P^E 2 =0,
( 2M ,« P _3Vi m ,)£ + 3vi m ,a
+ (p 2 ~1 ^2) E x + ? m2 E 2 = 0,
3\/3 Xi , /, u i/s D , 3\/3 \X 2
- -r Wl v + \ 2M p + — mi ) t
+ \ rn 1 E 1 + (p 2 - | m,j E 2 = 0.
(12-153)
The secular determinant is
(pi-nf + lm,) (-?»,) _( 2 M^P + ^ m2 ) (2^1 OT2 )
( 2M -p_3V| mj ) (3vf m2 ) ( p2 _| m2 ) (»„,)
-(?£..,) (»*"'* + ?£.,) (I-,) (P--|«,)
(12-154)
We know from previous considerations that this must be a fourth-degree
equation in P 2 , one root of which is P 2 = 0. This fact, together with the
symmetries in the array of coefficients, encourages us to try to manipulate
the above determinant to simplify its expansion and to bring out ex-
plicitly the factor P 2 . We add the second column to the first, and the
third to the fourth, then subtract the first row from the second, and the
third from the fourth :
= 0.
12-8]
STABILITY OF LAGRANGE S THREE BODIES
507
(P 2 -3M) (-i m ») -(21f" ! P + ^m ! ) -W U2 P)
[p 2 - 3M + | (m, + m 2 )l Um 1I2 P -^(m,- m 2 )]
[2M" 2 P + ^(«n- m,)] - [p 2 - j (m, + m,)]
= 0.
(12-155)
We factor P from the last column, then multiply the last column by
3M 1/2 /2, and subtract from the first column. We can then factor P from
the first column, to obtain
p (-\ m *) ~ (™ ll2 P + ^ m.) -(2M 1 ' 2 )
[p 2 - 3M + | (», + m,)l [~2M 1/2 P - ^0- (m, - m a )]
(1m-) (•£„..) (p 2 -^ 2 )
[2Af " 2 P + ^(»,- m a )] - [p 2 - j (mi + m,)]
(12-156)
The factor P 2 is now in evidence. To simplify the expansion of the de-
terminant, we multiply the third row by 2PM~ 112 and subtract from
the first row:
= 0.
* * -(2M 112 + 2P 2 M-" 2 )
[p 2 - ZM + I (m, + m„)] [w lla P - ^ (m, - m a )]
i^ 1
[2M" 2 P + ^ («., - m,)l - [p 2 - I (m, + m,)]
= 0.
(12-157)
The stars indicate terms that we do not write here, since they will not
appear in the result. We can now expand in minors of the first column,
and expand the resulting three-rowed determinant in minors of the last
column, with the final result:
P 2 (M + P 2 )^ + MP 2 + ^•(wiim 2 + m 2 m 3 + m a mi)] = 0.
(12-158)
Note, as a check on the algebra, that the three masses enter symmetrically
508 THEORY OF SMALL VIBRATIONS [CHAP. 12
in this equation, as they must. It is a fortunate accident that an addi-
tional factor appears explicitly, so that we have only to solve a quadratic
equation for P 2 . We have, finally, four roots:
p2 _ n P 2 = — M
^ ", r m, (12-159)
p 2 = —\M ± \\M 2 — 27(m 1 m 2 + m 2 m 3 + m 3 mi)] 1/2 .
As we know, the zero root results from the fact that there is an addi-
tional ignorable coordinate. The root P 2 = —M yields a stable oscilla-
tory mode. The last two roots for P 2 will both be real and negative
provided that
(mi + m 2 + m 3 ) 2 > 27(raim 2 + m 2 m 3 + m 3 m{). (12-160)
If this inequality is reversed, the last two roots are complex, and we have
four complex values of P. Of these roots two give rise to damped and two
to antidamped oscillatory solutions. In the intermediate case when the
two members of the inequality (12-160) are equal, it can be shown
that the amplitude of oscillation grows linearly in time. Hence the La-
grangian motion of three bodies is unstable when condition (12-160) is
not satisfied. If one of the bodies, say mi, is much smaller than the other
two, we have the restricted problem of three bodies studied in Section
7-6, and the condition for stability reduces to
(m 2 + m 3 ) 2 > 27m 2 m 3 , (12-161)
or, if m 3 is the largest,
m 3 > 24.96m 2 . (12-162)
If m 3 is the sun and m 2 is the planet Jupiter, this condition is satisfied;
hence, if we neglect the effects of all other planets, there are stable steady
motions in which a small body revolves around the sun with the same
period as Jupiter and at the corner of an equilateral triangle relative to
the sun and Jupiter. (There are two such positions.) The Trojan aster-
oids are a group of bodies with the same period as Jupiter which appear
to be in this position. Since Eq. (12-161) is also satisfied by the earth-
moon system, the corresponding steady motion of an artificial satellite
in the earth-moon system is stable. Consideration of motions perpendic-
ular to the plane of steady motion does not alter these conclusions. How-
ever, in view of the remarks at the end of Section 12-6, our conclusions
about the stability are valid only for limited periods of time.
We leave as an exercise the solution of Eqs. (12-153), to determine
the ratios of the variables and hence the oscillation pattern for each
normal mode (see Problem 30). In solving Eqs. (12-153), it may be
PROBLEMS 509
helpful to subject them to the same series of manipulations which led
from the determinant (12-154) to the determinant (12-157). Note that
adding one row of the determinant to another corresponds to adding the
corresponding equations. Adding two columns corresponds to grouping
the corresponding variables, that is, to introducing a new variable which
is a linear combination of the two original ones.
Problems
1. Find the transformation to normal coordinates for the two coupled oscil-
lators shown in Fig. 4-10.
2. Solve Problem 26, Chapter 4, by transforming from x\, X2 to normal co-
ordinates by the method of Section 12-3.
3. A mass m moving in space is subject to a force whose potential energy is
V = Vo exp [(5x 2 + 5y 2 + 8z 2 - 8yz - 2Qya - 8za)/a 2 ],
where the constants Vo and a are positive. Show that V has one minimum
point. Find the normal frequencies of vibration about the minimum.
4. A mass m is hung from a fixed support by a spring of constant k whose
relaxed length is I = 2mg/k. A second equal mass is hung from the first mass
by an identical spring. Find the six normal coordinates and the corresponding
frequencies for small vibrations of this system from its equilibrium position.
Each spring exerts a force only along the line joining its two ends, but may
pivot freely in any direction at its ends.
5. An ion of mass m, charge q, is held by a linear attractive force F = — kr
to a point A, where r is the distance from the ion to the point A. An identical
ion is similarly bound to a second point B a distance I from A. The two ions
move (in three-dimensional space) under the action of these forces and their
mutual electrostatic repulsion. Find the normal modes of vibration, and write
down the most general solution for small vibrations about the equilibrium point.
6. The mass W2 in Fig. 4-10 is subject to a force F2 = B sin ut. The sys-
tem is at rest at t = 0. Find the motion by the method of normal coordinates,
using the result of Problem 1 .
7. The pair of ions in Problem 5 is subject to a plane polarized electromagnetic
wave incident perpendicular to the line AB, whose electric field Eo cos cot is
directed at 45° to the line IB. Find the steady-state motion.
8. The mass in Problem 3 is subject to a force
F x = F y = F s = Be'"'.
Find a particular solution.
9. Set up the tensors M, B, K for Problem 25, Chapter 4, show that all
three can be simultaneously diagonalized, and solve the problem by the method
of normal coordinates.
10. The masses wii and rti2 in Fig. 4-10 are subject to frictional forces
— Tmixi, — TTO2X2, respectively. Find the general solution.
510 THEORY OF SMALL VIBRATIONS [CHAP. 12
11. Assume that V°(xi, . . . , x/) has a minimum at x\ = • • • = x; = 0, and
that V'(xi, . . . ,Xf) is small, but that Eq. (12-66) does not necessarily hold.
Find approximate expressions to first order in V and its derivatives at x\ = • • •
= Xf = 0, for the coordinates x\, . . . , x° f of the new equilibrium point for
V = V° +• V. If the expansion of V° about x\ = • • • = x/ = is given by
Eq. (12-67) (plus higher-order terms), and if the quadratic terms in the expan-
sion of V are to be
V = ]T i(K° kl + K' kl )y k y h
k,l
where y k = x k — x° k , find approximate first order expressions for the coeffi-
cients K' kl .
12. Find the second-order approximations for the coefficients Cj, C[, which
are given to first order by Eqs. (12-80) and (12-83).
13. Find the third-order approximation to the frequency correction given to
second order by Eq. (12-91).
14. Formulate the equations to be solved to obtain a first-order approxima-
tion in the case when conditions (12-87) fail for a group of four nearby modes,
i.e. the case of approximate degeneracy.
15. A triple pendulum is formed by suspending a mass M by a string of
length I from a fixed support. A mass m is hung from M by a string of length I,
and from this second mass a third mass m is hung by a third string of length I.
The masses swing in a single vertical plane. Set up the equations for small
vibrations of the system, using as coordinates the angles 6%, 02, 03 made by each
string with the vertical. Show that if M )$$> m, the normal coordinates can be
found if terms of order (m/M) 112 are neglected. Find the approximate normal
frequencies to order m/M. [Hint: Transform K to a constant tensor, and diag-
onalize M.]
16. In Fig. 12-3, the four masses move only along a horizontal straight line
under the action of four identical springs of constant k, and a weak spring of
constant k' <<C k. Find, to first order in k', an approximate solution for the
normal modes of vibration.
17. Find the approximate solution to Problem 16 for the case when the
masses are all equal. Does the approximate result suggest a way to solve the
problem exactly?
18. A uniform ellipsoid of revolution of mass M, whose axis of symmetry is
two thirds as long as its equatorial diameter, is modified by placing masses m,
2m, 3m, m, 2m, 3m in sequence around its equator at points 60° apart. Two
masses, each 4m, are placed at opposite ends of a diameter, making an angle of
45° with the axis and at the longitude of the masses m. If m <£. M, find, to
first order in m/M, the new principal axes. [Hint: The perturbation procedure
k m 1 k m2 k' m^ k m 4 k
Fig. 12-3. Four coupled harmonic oscillators.
PROBLEMS 511
developed in Section 12-5 for diagonalizing the tensor W may be applied to
diagonalize approximately any symmetric tensor 1° + I' if the eigenvectors of 1°
are known and I' is small.]
19. Apply the perturbation method to the problem of the string with variable
density considered in the last paragraph of Section 9-9, assuming that a <JC <ro-
Find the lowest normal frequency to second order in o, and write out the cor-
responding solution u(x, t) to first order in a.
20. Formulate a first-order perturbation method of solving Eqs. (12-60),
treating the friction as a small perturbation, and assuming the solution without
friction is already known. Show why, even in first order, one cannot introduce
normal coordinates which include the effects of friction.
21. Two charges -\-Ze are. located at fixed points a distance 2a apart. An
electron of mass m, charge — e, moves in the field of these charges. Find the
steady motions and the small vibrations about the steady motions.
*22. Two charges -{-Ze and — Ze are located at the fixed points z = a and
z = —a. An electron of mass m, charge — e, moves in the field of these charges.
Sketch a graph of z vs. r, where r is the distance from the z-axis, showing the
values of z, r for which there are steady motions. Investigate the stability of
these steady motions.
23. A mass m slides without friction on a smooth horizontal table. It is tied
to a weightless string of total length I which passes through a hole in the table
and is tied at its lower end to a mass M which hangs below the table. Set up
the Hamiltonian function using as coordinates the polar coordinates r, a of the
mass m relative to the hole, and the spherical angles 9, <p of the mass M relative
to the hole. Find the steady motions and the normal frequencies of small vibra-
tions about a steady motion.
24. Assume that the three masses in Fig. 4-16 are free to move in a plane but
are constrained to remain in a straight line relative to one another. Choose
your coordinates so that as many as possible will be ignorable, find the steady
motions, and find the normal modes of vibration about them.
25. A symmetrical rigid body is mounted in weightless, frictionless gimbal
rings. A hairspring is attached to one of the rings so as to exert a restoring
torque — h}> about the z-axis, where <t> is the Euler angle. Find the steady mo-
tions and investigate the character of small vibrations about them.
26. In Problem 13, Chapter 11, a hairspring is connected between the disk
axle and the rings which exerts a restoring torque — b//', where \[>' is the rela-
tive angle of rotation between disk and rings. The "gyroscope" moves freely
in space with no external forces. Find the steady motions and investigate the
small vibrations about them.
27. Two masses m are connected by a rigid weightless rod of length 21. One
mass is connected with the origin by a spring of constant k, the other by a
spring of constant 2k. The relaxed length of both springs is zero. The masses
move in a single plane. Choose as coordinates the polar coordinates r, 6 of the
center of mass relative to the origin, and the angle a which the rod makes with
the radius from the origin to the center of mass, taking a = when the stronger
spring is stretched least. Find the steady motions and the conditions under
which they are stable.
512 THEORY OF SMALL VIBRATIONS [CHAP. 12
*28. In Problem 23, a second mass m slides without friction on the table, and
is connected to the first mass by a rigid weightless rod of length a. (Assume
the arrangement is ingeniously contrived so that the rod and string do not
become entangled.) Assume also that M = 2m. Use as an additional coordi-
nate the angle /? between the rod and the string. Find the steady motions, and
determine which are stable. Find the normal vibrations about the stable steady
motions.
29. The vector potential due to a magnetic dipole of magnetic moment n is,
in spherical coordinates relative to the dipole axis,
H sin 6
A = ffl.
2*r 2
Find the steady motions for a charged particle moving in such a field, and
show that they are unstable.
30. Find the solution of Eqs. (12-153) for Xi, X 2 , Ei, E 2 , when P 2 = —M,
and describe the corresponding oscillation. [See the hint in the last paragraph
of Section 12-8.]
*31. Analyze the case which was omitted in Section 12-8 when the three
bodies mi, m 2 , mz lie in a straight line. Show that there are three possible
steady motions, one for each mass lying between the other two. [Hint: You
will need Descartes' rule of signs.] Show that motions near each of these steady
motions are unstable. Compare your results with those of Section 7-6 and
Problem 17 of Chapter 7.
32. Find the solution of Eqs. (12-153) for the double root P 2 = —ffl,
when the inequality (12-160) becomes an equality. Show that in this case,
Eqs. (12-150) have a second solution in which Xi, X 2 , E\, E 2 are certain linear
functions of t, say Xi = X{ + Xi't, etc. (You can simplify the algebra a little
by assuming that one of the additive constants, say X[, is zero. This is allow-
able, since X{ can always be made zero by subtracting from the second solution
a suitable multiple of the first solution you found in which Xi is constant.
The linearity of the equations permits linear superposition of solutions.)
33. Find the solution of Eqs. (12-153) for X h X 2 , E lt E 2 , for. the root P 2 = 0,
and show that it corresponds to a new steady motion near the chosen one.
Since your solution has only one arbitrary constant, there must be another
solution of Eqs. (12-150) corresponding to P 2 = 0. Guess its form, and verify
by substitution.
*34. Show that if motions of Lagrange's three bodies out of the plane of the
steady motion are considered, at least one of the three additional coordinates
is ignorable. Choose as two nonignorable coordinates the distances qi = z\ — zz
and q 2 = z 2 — 23, where z% is the perpendicular distance of to,- from